代码随想录Day_01|Leetcode.704.35.34.27（二分查找）

704.二分查找

[low,high]：当target小于index，high赋值index-1，将这次的index排除在外，这样下次的判断范围就是low~high，当high等于low时判断最后一个元素。

public int search(int[] nums, int target) {
        int high = nums.length - 1;
        int low = 0;
        while (high >= low) {
            int index = (high - low) / 2 + low;
            if (nums[index] > target)
                high = index - 1;
            else if (nums[index]<target) {
                low = index + 1;
            }else{
                return index;
            }
        }
        return -1;
    }

[low,high)：当target小于index，high赋值index而不是index-1，这样下次判断范围就是low~high-1，所以high必须大于low，当high等于low时循环结束。

   public int search(int[] nums, int target) {
        int high = nums.length ;
        int low = 0;
        while (high > low) {
            int index = (high - low) / 2 + low;
            if (nums[index] > target)
                high = index;
            else if (nums[index] < target) {
                low = index + 1;
            } else {
                return index;
            }
        }
        return -1;
    }

 35.搜索插入位置

[low,high]：当不在数组中，循环结束时high=low-1，target一定大于high小于low，所以插入位置为low

public int searchInsert(int[] nums, int target) {
        int high = nums.length - 1;
        int low = 0;
        while (high >= low) {
            int index = (high - low) / 2 + low;
            if (nums[index] > target)
                high = index - 1;
            else if (nums[index] < target) {
                low = index + 1;
            } else {
                return index;
            }
        }
        return low;
    }

34.排序数组查找第一个和最后一个元素

先写一种解法，首先找到一个在目标区间的值的索引，然后分别寻找左右区间，寻找时候记得在判断条件先判断是否超出索引，在判断目标是否为target，否则会报错。这是因为&&会从左到右的顺序依次判断条件。

i >= 0 && nums[i] == target
j <= nums.length - 1 && nums[j] == target

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = {-1, -1};
        int index = searchmiddle(nums, target);
        if (index == -1)
            return result;
        int j = index;
        int i = index;
        while (i >= 0 && nums[i] == target) {
            i -= 1;
        }
        while (j <= nums.length - 1 && nums[j] == target) {
            j += 1;
        }
        result[0] = i + 1;
        result[1] = j - 1;
        return result;
    }

    public int searchmiddle(int[] nums, int target) {
        int high = nums.length - 1;
        int low = 0;
        while (high >= low) {
            int index = (high - low) / 2 + low;
            if (nums[index] > target)
                high = index - 1;
            else if (nums[index] < target) {
                low = index + 1;
            } else {
                return index;
            }
        }
        return -1;
    }
}

第二种解法考虑三种情况：

• 情况一：target 在数组范围的右边或者左边，例如数组{3, 4, 5}，target为2或者数组{3, 4, 5},target为6，此时应该返回{-1, -1}
• 情况二：target 在数组范围中，且数组中不存在target，例如数组{3,6,7},target为5，此时应该返回{-1, -1}
• 情况三：target 在数组范围中，且数组中存在target，例如数组{3,6,7},target为6，此时应该返回{1, 1}

 三种情况的解法没有想到，后期会回顾。

class Solution {
    int[] searchRange(int[] nums, int target) {
        int leftBorder = getLeftBorder(nums, target);
        int rightBorder = getRightBorder(nums, target);
        // 情况一
        if (leftBorder == -2 || rightBorder == -2) return new int[]{-1, -1};
        // 情况三
        if (rightBorder - leftBorder > 1) return new int[]{leftBorder + 1, rightBorder - 1};
        // 情况二
        return new int[]{-1, -1};
    }

    int getRightBorder(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int rightBorder = -2; // 记录一下rightBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] > target) {
                right = middle - 1;
            } else { // 寻找右边界，nums[middle] == target的时候更新left
                left = middle + 1;
                rightBorder = left;
            }
        }
        return rightBorder;
    }

    int getLeftBorder(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int leftBorder = -2; // 记录一下leftBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] >= target) { // 寻找左边界，nums[middle] == target的时候更新right
                right = middle - 1;
                leftBorder = right;
            } else {
                left = middle + 1;
            }
        }
        return leftBorder;
    }
}

27.移除元素

快慢指针

class Solution {
    public int removeElement(int[] nums, int val) {
        int slow = 0;
        int fast = 0;
        while (fast <= nums.length - 1) {
            if (nums[fast] != val) {
                nums[slow] = nums[fast];
                slow += 1;
            }
            fast += 1;
        }
    return slow;
    }
}

双指针

public int removeElement(int[] nums, int val) {
        int left = 0;
        int right = nums.length - 1;
        while(left <= right){
            if(nums[left] == val){
                nums[left] = nums[right];
                right--;
            }else {
                // 这里兼容了right指针指向的值与val相等的情况
                left++;
            }
        }
        return left;
    }