双指针代码

#双指针
移动0到右边
全局+部分
复杂度
n和1
输入: nums = [0,1,0,3,12]
输出: [1,3,12,0,0]

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        n = len(nums)
        left = right = 0
        while right < n:
            if nums[right] != 0:
                nums[left], nums[right] = nums[right], nums[left]
                left += 1
            right += 1

func moveZeroes(nums []int) {
    left, right, n := 0, 0, len(nums)
    for right < n {
        if nums[right] != 0 {
            nums[left], nums[right] = nums[right], nums[left]
            left++
        }
        right++
    }
}

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int n = nums.size(), left = 0, right = 0;
        while (right < n) {
            if (nums[right]) {
                swap(nums[left], nums[right]);
                left++;
            }
            right++;
        }
    }
};

class Solution {
    public void moveZeroes(int[] nums) {
        int n = nums.length, left = 0, right = 0;
        while (right < n) {
            if (nums[right] != 0) {
                swap(nums, left, right);
                left++;
            }
            right++;
        }
    }

    public void swap(int[] nums, int left, int right) {
        int temp = nums[left];
        nums[left] = nums[right];
        nums[right] = temp;
    }
}

void swap(int *a, int *b) {
    int t = *a;
    *a = *b, *b = t;
}

void moveZeroes(int *nums, int numsSize) {
    int left = 0, right = 0;
    while (right < numsSize) {
        if (nums[right]) {
            swap(nums + left, nums + right);
            left++;
        }
        right++;
    }
}

#给定一个长度为 n 的整数数组 height 。有 n 条垂线，第 i 条线的两个端点是 (i, 0) 和 (i, height[i]) 。

class Solution:
    def maxArea(self, height: List[int]) -> int:
        l, r = 0, len(height) - 1
        ans = 0
        while l < r:
            area = min(height[l], height[r]) * (r - l)
            ans = max(ans, area)
            if height[l] <= height[r]:
                l += 1
            else:
                r -= 1
        return ans

public class Solution {
    public int maxArea(int[] height) {
        int l = 0, r = height.length - 1;
        int ans = 0;
        while (l < r) {
            int area = Math.min(height[l], height[r]) * (r - l);
            ans = Math.max(ans, area);
            if (height[l] <= height[r]) {
                ++l;
            }
            else {
                --r;
            }
        }
        return ans;
    }
}

class Solution {
public:
    int maxArea(vector<int>& height) {
        int l = 0, r = height.size() - 1;
        int ans = 0;
        while (l < r) {
            int area = min(height[l], height[r]) * (r - l);
            ans = max(ans, area);
            if (height[l] <= height[r]) {
                ++l;
            }
            else {
                --r;
            }
        }
        return ans;
    }
};

三元数三个加块为0
排序加双指针时间复杂度：
O(N 2)，其中 N 是数组 nums 的长度。

空间复杂度：O(logN)。我们忽略存储答案的空间，额外的排序的空间复杂度为 O(logN)。然而我们修改了输入的数组 nums，在实际情况下不一定允许，因此也可以看成使用了一个额外的数组存储了 nums 的副本并进行排序，空间复杂度为 O(N)。
输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        nums.sort()
        ans = list()
        
        # 枚举 a
        for first in range(n):
            # 需要和上一次枚举的数不相同
            if first > 0 and nums[first] == nums[first - 1]:
                continue
            # c 对应的指针初始指向数组的最右端
            third = n - 1
            target = -nums[first]
            # 枚举 b
            for second in range(first + 1, n):
                # 需要和上一次枚举的数不相同
                if second > first + 1 and nums[second] == nums[second - 1]:
                    continue
                # 需要保证 b 的指针在 c 的指针的左侧
                while second < third and nums[second] + nums[third] > target:
                    third -= 1
                # 如果指针重合，随着 b 后续的增加
                # 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
                if second == third:
                    break
                if nums[second] + nums[third] == target:
                    ans.append([nums[first], nums[second], nums[third]])
        
        return ans

func threeSum(nums []int) [][]int {
    n := len(nums)
    sort.Ints(nums)
    ans := make([][]int, 0)
 
    // 枚举 a
    for first := 0; first < n; first++ {
        // 需要和上一次枚举的数不相同
        if first > 0 && nums[first] == nums[first - 1] {
            continue
        }
        // c 对应的指针初始指向数组的最右端
        third := n - 1
        target := -1 * nums[first]
        // 枚举 b
        for second := first + 1; second < n; second++ {
            // 需要和上一次枚举的数不相同
            if second > first + 1 && nums[second] == nums[second - 1] {
                continue
            }
            // 需要保证 b 的指针在 c 的指针的左侧
            for second < third && nums[second] + nums[third] > target {
                third--
            }
            // 如果指针重合，随着 b 后续的增加
            // 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
            if second == third {
                break
            }
            if nums[second] + nums[third] == target {
                ans = append(ans, []int{nums[first], nums[second], nums[third]})
            }
        }
    }
    return ans
}

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        // 枚举 a
        for (int first = 0; first < n; ++first) {
            // 需要和上一次枚举的数不相同
            if (first > 0 && nums[first] == nums[first - 1]) {
                continue;
            }
            // c 对应的指针初始指向数组的最右端
            int third = n - 1;
            int target = -nums[first];
            // 枚举 b
            for (int second = first + 1; second < n; ++second) {
                // 需要和上一次枚举的数不相同
                if (second > first + 1 && nums[second] == nums[second - 1]) {
                    continue;
                }
                // 需要保证 b 的指针在 c 的指针的左侧
                while (second < third && nums[second] + nums[third] > target) {
                    --third;
                }
                // 如果指针重合，随着 b 后续的增加
                // 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
                if (second == third) {
                    break;
                }
                if (nums[second] + nums[third] == target) {
                    ans.push_back({nums[first], nums[second], nums[third]});
                }
            }
        }
        return ans;
    }
};

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        // 枚举 a
        for (int first = 0; first < n; ++first) {
            // 需要和上一次枚举的数不相同
            if (first > 0 && nums[first] == nums[first - 1]) {
                continue;
            }
            // c 对应的指针初始指向数组的最右端
            int third = n - 1;
            int target = -nums[first];
            // 枚举 b
            for (int second = first + 1; second < n; ++second) {
                // 需要和上一次枚举的数不相同
                if (second > first + 1 && nums[second] == nums[second - 1]) {
                    continue;
                }
                // 需要保证 b 的指针在 c 的指针的左侧
                while (second < third && nums[second] + nums[third] > target) {
                    --third;
                }
                // 如果指针重合，随着 b 后续的增加
                // 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
                if (second == third) {
                    break;
                }
                if (nums[second] + nums[third] == target) {
                    List<Integer> list = new ArrayList<Integer>();
                    list.add(nums[first]);
                    list.add(nums[second]);
                    list.add(nums[third]);
                    ans.add(list);
                }
            }
        }
        return ans;
    }
}

接雨水
动态规划 单调栈 双指针

n和n

//python
class Solution:
    def trap(self, height: List[int]) -> int:
        if not height:
            return 0
        
        n = len(height)
        leftMax = [height[0]] + [0] * (n - 1)
        for i in range(1, n):
            leftMax[i] = max(leftMax[i - 1], height[i])

        rightMax = [0] * (n - 1) + [height[n - 1]]
        for i in range(n - 2, -1, -1):
            rightMax[i] = max(rightMax[i + 1], height[i])

        ans = sum(min(leftMax[i], rightMax[i]) - height[i] for i in range(n))
        return ans

//cpp
class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        if (n == 0) {
            return 0;
        }
        vector<int> leftMax(n);
        leftMax[0] = height[0];
        for (int i = 1; i < n; ++i) {
            leftMax[i] = max(leftMax[i - 1], height[i]);
        }

        vector<int> rightMax(n);
        rightMax[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            rightMax[i] = max(rightMax[i + 1], height[i]);
        }

        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += min(leftMax[i], rightMax[i]) - height[i];
        }
        return ans;
    }
};

//c
int trap(int* height, int heightSize) {
    int n = heightSize;
    if (n == 0) {
        return 0;
    }
    int leftMax[n];
    memset(leftMax, 0, sizeof(leftMax));
    leftMax[0] = height[0];
    for (int i = 1; i < n; ++i) {
        leftMax[i] = fmax(leftMax[i - 1], height[i]);
    }

    int rightMax[n];
    memset(rightMax, 0, sizeof(rightMax));
    rightMax[n - 1] = height[n - 1];
    for (int i = n - 2; i >= 0; --i) {
        rightMax[i] = fmax(rightMax[i + 1], height[i]);
    }

    int ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += fmin(leftMax[i], rightMax[i]) - height[i];
    }
    return ans;
}

//go
func trap(height []int) (ans int) {
    n := len(height)
    if n == 0 {
        return
    }

    leftMax := make([]int, n)
    leftMax[0] = height[0]
    for i := 1; i < n; i++ {
        leftMax[i] = max(leftMax[i-1], height[i])
    }

    rightMax := make([]int, n)
    rightMax[n-1] = height[n-1]
    for i := n - 2; i >= 0; i-- {
        rightMax[i] = max(rightMax[i+1], height[i])
    }

    for i, h := range height {
        ans += min(leftMax[i], rightMax[i]) - h
    }
    return
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

//java
class Solution {
    public int trap(int[] height) {
        int n = height.length;
        if (n == 0) {
            return 0;
        }

        int[] leftMax = new int[n];
        leftMax[0] = height[0];
        for (int i = 1; i < n; ++i) {
            leftMax[i] = Math.max(leftMax[i - 1], height[i]);
        }

        int[] rightMax = new int[n];
        rightMax[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            rightMax[i] = Math.max(rightMax[i + 1], height[i]);
        }

        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += Math.min(leftMax[i], rightMax[i]) - height[i];
        }
        return ans;
    }
}

//javascript
var trap = function(height) {
    const n = height.length;
    if (n == 0) {
        return 0;
    }

    const leftMax = new Array(n).fill(0);
    leftMax[0] = height[0];
    for (let i = 1; i < n; ++i) {
        leftMax[i] = Math.max(leftMax[i - 1], height[i]);
    }

    const rightMax = new Array(n).fill(0);
    rightMax[n - 1] = height[n - 1];
    for (let i = n - 2; i >= 0; --i) {
        rightMax[i] = Math.max(rightMax[i + 1], height[i]);
    }

    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += Math.min(leftMax[i], rightMax[i]) - height[i];
    }
    return ans;
};

n和n

//python
class Solution:
    def trap(self, height: List[int]) -> int:
        ans = 0
        stack = list()
        n = len(height)
        
        for i, h in enumerate(height):
            while stack and h > height[stack[-1]]:
                top = stack.pop()
                if not stack:
                    break
                left = stack[-1]
                currWidth = i - left - 1
                currHeight = min(height[left], height[i]) - height[top]
                ans += currWidth * currHeight
            stack.append(i)
        
        return ans

//cpp
class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        stack<int> stk;
        int n = height.size();
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && height[i] > height[stk.top()]) {
                int top = stk.top();
                stk.pop();
                if (stk.empty()) {
                    break;
                }
                int left = stk.top();
                int currWidth = i - left - 1;
                int currHeight = min(height[left], height[i]) - height[top];
                ans += currWidth * currHeight;
            }
            stk.push(i);
        }
        return ans;
    }
};

//c
int trap(int* height, int heightSize) {
    int n = heightSize;
    if (n == 0) {
        return 0;
    }
    int ans = 0;
    int stk[n], top = 0;
    for (int i = 0; i < n; ++i) {
        while (top && height[i] > height[stk[top - 1]]) {
            int stk_top = stk[--top];
            if (!top) {
                break;
            }
            int left = stk[top - 1];
            int currWidth = i - left - 1;
            int currHeight = fmin(height[left], height[i]) - height[stk_top];
            ans += currWidth * currHeight;
        }
        stk[top++] = i;
    }
    return ans;
}

//go
func trap(height []int) (ans int) {
    stack := []int{}
    for i, h := range height {
        for len(stack) > 0 && h > height[stack[len(stack)-1]] {
            top := stack[len(stack)-1]
            stack = stack[:len(stack)-1]
            if len(stack) == 0 {
                break
            }
            left := stack[len(stack)-1]
            curWidth := i - left - 1
            curHeight := min(height[left], h) - height[top]
            ans += curWidth * curHeight
        }
        stack = append(stack, i)
    }
    return
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

//java
class Solution {
    public int trap(int[] height) {
        int ans = 0;
        Deque<Integer> stack = new LinkedList<Integer>();
        int n = height.length;
        for (int i = 0; i < n; ++i) {
            while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
                int top = stack.pop();
                if (stack.isEmpty()) {
                    break;
                }
                int left = stack.peek();
                int currWidth = i - left - 1;
                int currHeight = Math.min(height[left], height[i]) - height[top];
                ans += currWidth * currHeight;
            }
            stack.push(i);
        }
        return ans;
    }
}

//javascript
var trap = function(height) {
    let ans = 0;
    const stack = [];
    const n = height.length;
    for (let i = 0; i < n; ++i) {
        while (stack.length && height[i] > height[stack[stack.length - 1]]) {
            const top = stack.pop();
            if (!stack.length) {
                break;
            }
            const left = stack[stack.length - 1];
            const currWidth = i - left - 1;
            const currHeight = Math.min(height[left], height[i]) - height[top];
            ans += currWidth * currHeight;
        }
        stack.push(i);
    }
    return ans;
};

双指针
n和1

class Solution:
    def trap(self, height: List[int]) -> int:
        ans = 0
        left, right = 0, len(height) - 1
        leftMax = rightMax = 0

        while left < right:
            leftMax = max(leftMax, height[left])
            rightMax = max(rightMax, height[right])
            if height[left] < height[right]:
                ans += leftMax - height[left]
                left += 1
            else:
                ans += rightMax - height[right]
                right -= 1
        
        return ans

func trap(height []int) (ans int) {
    left, right := 0, len(height)-1
    leftMax, rightMax := 0, 0
    for left < right {
        leftMax = max(leftMax, height[left])
        rightMax = max(rightMax, height[right])
        if height[left] < height[right] {
            ans += leftMax - height[left]
            left++
        } else {
            ans += rightMax - height[right]
            right--
        }
    }
    return
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        int left = 0, right = height.size() - 1;
        int leftMax = 0, rightMax = 0;
        while (left < right) {
            leftMax = max(leftMax, height[left]);
            rightMax = max(rightMax, height[right]);
            if (height[left] < height[right]) {
                ans += leftMax - height[left];
                ++left;
            } else {
                ans += rightMax - height[right];
                --right;
            }
        }
        return ans;
    }
};

int trap(int* height, int heightSize) {
    int ans = 0;
    int left = 0, right = heightSize - 1;
    int leftMax = 0, rightMax = 0;
    while (left < right) {
        leftMax = fmax(leftMax, height[left]);
        rightMax = fmax(rightMax, height[right]);
        if (height[left] < height[right]) {
            ans += leftMax - height[left];
            ++left;
        } else {
            ans += rightMax - height[right];
            --right;
        }
    }
    return ans;
}

class Solution {
    public int trap(int[] height) {
        int ans = 0;
        int left = 0, right = height.length - 1;
        int leftMax = 0, rightMax = 0;
        while (left < right) {
            leftMax = Math.max(leftMax, height[left]);
            rightMax = Math.max(rightMax, height[right]);
            if (height[left] < height[right]) {
                ans += leftMax - height[left];
                ++left;
            } else {
                ans += rightMax - height[right];
                --right;
            }
        }
        return ans;
    }
}

var trap = function(height) {
    let ans = 0;
    let left = 0, right = height.length - 1;
    let leftMax = 0, rightMax = 0;
    while (left < right) {
        leftMax = Math.max(leftMax, height[left]);
        rightMax = Math.max(rightMax, height[right]);
        if (height[left] < height[right]) {
            ans += leftMax - height[left];
            ++left;
        } else {
            ans += rightMax - height[right];
            --right;
        }
    }
    return ans;
};