数组类算法 - 合集_line 1122: char 34: runtime error: addition of uns-优快云博客

This topic likes delete the repeat. As the topic is double pointer, so use two pointer here. The first pointer moves faster so just name it fast. The second pointer moves slower and name it slow. The most significant code is

if (nums[fast + 1] != nums[fast])
{
nums[slow] = nums[fast];
}

and the code comes easy:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        
        int slow = 1; // 慢指针从索引1开始
        for (int fast = 1; fast < n; fast++) { // 快指针从索引1开始
            if (nums[fast] != nums[fast - 1]) {
                nums[slow] = nums[fast]; // 复制不同的元素
                slow++; // 慢指针向前移动
            }
        }
      
        return slow; // 返回新的数组长度
    }
};

*************

Guess what, delete repeat ones if it repeat 3 or more times：

80. 删除有序数组中的重复项 II - 力扣（LeetCode）https://leetcode.cn/problems/remove-duplicates-from-sorted-array-ii/description/

It does have some challenge. The nums is orsered and it makes a little bit easy. Compare the nums[fast + 2] with nums[fast]. It ensures only 2 elements remains the same. If you want keep 3 same elements, just compare nums[fast + 3] with nums[fast].

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {

        
        int n = nums.size();

        int slow = 0; // 慢指针，指向下一个可能的非重复元素的位置
        for (int fast = 2; fast < n; fast++) { // 快指针，遍历数组
            if (nums[fast] != nums[slow + 2]) { // 如果当前元素不等于慢指针前两个位置的元素
                nums[slow] = nums[fast]; // 将当前元素复制到慢指针的位置，并移动慢指针
                slow++;
            }
        }

        return slow; // 返回新数组的长度
    }
};

The code dosenot run:

Line 1122: Char 34: runtime error: addition of unsigned offset to 0x503000000070 overflowed to 0x50300000006c (stl_vector.h)
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior /usr/lib/gcc/x86_64-linux-gnu/14/../../../../include/c++/14/bits/stl_vector.h:1131:34

This is because data goes beyond borders. For example:

nums = [1 2 3]

n = 2

if fast = 2

nums[2] != nums[4]

however nums[4] 实际是不存在的

Take borders into consideration:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() <= 2) {
            return nums.size(); // 如果数组长度小于等于2，不需要删除任何元素
        }

        int slow = 2; // 慢指针，指向下一个可能的非重复元素的位置
        for (int fast = 2; fast < nums.size(); fast++) { // 快指针，遍历数组
            if (nums[fast] != nums[slow - 2]) { // 如果当前元素不等于慢指针前两个位置的元素
                nums[slow] = nums[fast]; // 将当前元素复制到慢指针的位置，并移动慢指针
                slow++;
            }
        }

        return slow; // 返回新数组的长度
    }
};

*************

Nums are ordered in the topics before. And what if the order is unordered?

75. 颜色分类 - 力扣（LeetCode）

https://leetcode.cn/problems/sort-colors/description/

Need to order them without using sort. Let's see if I use sort:

One line code runs.

And use double pointer seems a little bit noncommittal. Only 3 elements: 0 1 2, just ordering 0 and 1 is fine. The rest of the elements will place themselves at the right place. Use double pointer. If nums[left] == 0, moves left. If nums[right] == 2, moves right.

class Solution {
public:
    void sortColors(vector<int>& nums) {

        int n = nums.size();
        int left = 0;
        int right = n - 1;
        int current = 0;

        for (current = 0; current < n; current++){
            if (nums[current] == 0){
                nums[left] = nums[current];
                left++;
            } else if (nums[current] == 2){
                nums[right] = nums[current];
                right--;
            } 
        }
    }
};

A brilliant idea whit running wrong:

Simulate the progress:

Initial array: [2, 0, 2, 1, 1, 0]
left = 0, right = 5, current = 0
Iteration 1:
current = 0, nums[0] = 2
Assign nums[5] = 2 (nums[5] was 0, now it's 2)
right = 4
Array: [2, 0, 2, 1, 1, 2]
Iteration 2:
current = 1, nums[1] = 0
Assign nums[0] = 0
left = 1
Array: [0, 0, 2, 1, 1, 2]
Iteration 3:
current = 2, nums[2] = 2
Assign nums[4] = 2 (nums[4] was 1, now it's 2)
right = 3
Array: [0, 0, 2, 1, 2, 2]
Iteration 4:
current = 3, nums[3] = 1
Do nothing
current = 4
Iteration 5:
current = 4, nums[4] = 2
Assign nums[3] = 2
right = 2
Array: [0, 0, 2, 2, 2, 2]
Iteration 6:
current = 5, nums[5] = 2
Assign nums[2] = 2
right = 1
Array: [0, 0, 2, 2, 2, 2]
End of loop.

The problem is that just swap.

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        int left = 0;
        int right = n - 1;
        int current = 0;

        while (current <= right) {
            if (nums[current] == 0) {
                swap(nums[left], nums[current]);
                left++;
                current++;
            } else if (nums[current] == 2) {
                swap(nums[right], nums[current]);
                right--;
            } else { // nums[current] == 1
                current++;
            }
        }
    }
};

*************

215. 数组中的第K个最大元素 - 力扣（LeetCode）

https://leetcode.cn/problems/kth-largest-element-in-an-array/description/

First eye on the topic, the demand confuse me. Note that it is the kth largest element in the sorted order, not the kth distinct element. The requirement is to return nums[n - k + 1].

First at the begining, bouble sort comes first. However sorting is forbidden.

Still cannot solve it. leave it alone.

*************

88. 合并两个有序数组 - 力扣（LeetCode）

https://leetcode.cn/problems/merge-sorted-array/

Merge two ordered arries. Copy nums1 and nums2 to make a new array 3 and sort array 3:

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        // 创建临时数组
        vector<int> temp(m + n);
        
        // 复制 nums1 的前 m 个元素到 temp
        for (int i = 0; i < m; i++) {
            temp[i] = nums1[i];
        }
        
        // 复制 nums2 的前 n 个元素到 temp
        for (int i = 0; i < n; i++) {
            temp[m + i] = nums2[i];
        }
        
        // 对 temp 进行排序
        sort(temp.begin(), temp.end());
        
        // 将排序后的 temp 复制回 nums1
        for (int i = 0; i < m + n; i++) {
            nums1[i] = temp[i];
        }
    }
};

*************

167. 两数之和 II - 输入有序数组 - 力扣（LeetCode）

https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/description/

Think about it. I have two pointers i and j. make pointer moves from left. Make pointer j moves from right. each moving estimating numbers[i[] + numsbers[j] != target.

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        
        int n = numbers.size();

        for (int i = 0; i < n; i++){
            for (int j = n - 1; j > 0; j--){
                if (numbers[i] + numbers[j] == target){
                    
                    return {i + 1, j + 1};
                }
            }
        }

        return { };
    }
};

It runs overtime. I cannot understand.

copy the answer:

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left = 0;
        int right = numbers.size() - 1;
        
        while (left < right) {
            int sum = numbers[left] + numbers[right];
            if (sum == target) {
                return {left + 1, right + 1};
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
        // 题目保证有解，所以理论上不会执行到这一步
        return {};
    }
};

*************

125. 验证回文串 - 力扣（LeetCode）

https://leetcode.cn/problems/valid-palindrome/description/

This topic is the last topic of this paper because I have do a lot about Longest Palindromic Substring.

In this topic, count the Longest Palindromic Substring of string:

最长回文子序列-优快云博客https://blog.youkuaiyun.com/ElseWhereR/article/details/143785381?spm=1001.2014.3001.5502

class Solution {
public:
    bool isPalindrome(string s) {
        
        int n = s.size(); // get the length of the string
        vector<vector<int>> dp(n, vector<int>(n, 0)); // initialise the array into elements zero
 
        for (int i = 0; i < n; i++){
            dp[i][i] = 1; // ez to understand that one letter is always true
        }
 
        // caculate the length more than 2
        for (int len = 2; len <= n; ++len) { // out-cicle is the length of the new string
            for (int i = 0; i <= n - len; ++i) { // in-cicle start from the different place
                int j = i + len - 1; // define the end of the string
                if (s[i] == s[j]) { // in new srtring, stare == end, which has the possibility, so need check
                    dp[i][j] = (i + 1 <= j - 1 ? dp[i + 1][j - 1] : 0) + 2;
                } else {
                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        
        // dp[0][n-1] return
        return dp[0][n - 1];

        if (dp[0][n - 1] = n){
            return true;
        }else{
            return false;
        }

    }
};