1、移动零

(1) 题目解析

(2) 算法原理

class Solution {
public:void moveZeroes(vector<int>& nums) {for(int cur=0,dest=-1;cur<nums.size();++cur){if(nums[cur]) swap(nums[cur],nums[++dest]);}}
};

2、复写零

(1) 题目解析

(2) 算法原理

class Solution {
public:void duplicateZeros(vector<int>& arr) {// 1.找到最后一个下标位置int cur = 0;int dest = -1;int n = arr.size();while(dest < n){if(arr[cur]) dest++;else dest += 2;// dest走到边界if(dest >= n-1) break;cur++;}// 处理边界存在最后一个数为0if(dest == n){arr[n-1] = 0;cur--,dest-=2;}while(cur >= 0){if(arr[cur])  arr[dest--] = arr[cur--];else{arr[dest--] = 0;arr[dest--] = 0;cur--;}}}
};

3、快乐数

(1) 题目解析

(2) 算法原理

class Solution {
public:int bitSum(int n){int sum = 0;while(n){int a = n % 10;sum += pow(a,2);n /= 10;}return sum;}bool isHappy(int n) {// 快慢指针int slow = n,fast = bitSum(n);while(slow != fast){slow = bitSum(slow);fast = bitSum(bitSum(fast));}return slow == 1 ? true : false;}
};

4、盛最多水的容器

(1) 题目解析

(2) 算法原理

class Solution {
public:int maxArea(vector<int>& arr) {int n = arr.size();int left=0,right=n-1;int ret = 0;while(left < right){int height = min(arr[right],arr[left]);int v = (right-left) * height;ret = max(ret,v);height == arr[right] ? right--:left++;} return ret;}
};

5、有效三角）——形个数

(1) 题目解析

(2) 算法原理

class Solution {
public:int triangleNumber(vector<int>& nums) {int n = nums.size();sort(nums.begin(),nums.end());int num = 0;for(int c=nums.size()-1;c>=2;--c){int left = 0;int right = c-1;while(left < right){int sum = nums[left] + nums[right];if(sum > nums[c]){num += (right-left);right--;}else left++;}}return num;}
};

6、和为s的两个数

(1) 题目解析

(2) 算法原理

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {int n = nums.size();int left = 0,right=n-1;while(left < right){int sum = nums[left]+ nums[right];if(sum > target){right--;}else if(sum < target){left++;}else return {nums[left],nums[right]};}return {-1};}
};

7、三数之和

(1) 题目解析

(2) 算法原理

class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {int n = nums.size();vector<vector<int>> res;sort(nums.begin(),nums.end());for(int i=0;i<n;){if(nums[i] > 0) break;int left = i+1,right = n-1;while(left < right){int tar = -nums[i];int sum = nums[left] + nums[right];if(sum > tar) right--;else if(sum < tar) left++;else{res.push_back({nums[i],nums[left],nums[right]});left++,right--;// 区间数比较while(left <right && nums[left] == nums[left-1]) left++;while(left < right && nums[right] == nums[right+1] ) right--;}}// 固定数比较i++;while(i < n && nums[i] == nums[i-1]) i++;}return res;}
};

8、四数之和

(1) 题目解析

这道题的本质和三数之和没什么区别，只不过比三数字和多套一层循环就能够解决了。所以不再细讲，注意到各个循环内的越界去重问题就行。

(2) 算法原理

class Solution {
public:vector<vector<int>> fourSum(vector<int>& nums, int target) {int n = nums.size();// 排序sort(nums.begin(),nums.end());vector<vector<int>> res;for(int i=0;i<n;){for(int j=i+1;j<n;){int left =j+1,right=n-1;while(left < right){// 固定两个数// 可能出现溢出情况long long tar = (long long)target - nums[i] - nums[j];int sum = nums[left] + nums[right];if(sum > tar) right--;else if(sum < tar) left++;else {res.push_back({nums[i],nums[j],nums[left],nums[right]});left++,right--;while(left < right && nums[left] == nums[left-1]) left++;while(left < right && nums[right] == nums[right+1]) right--;}}j++;while(j < n && nums[j] == nums[j-1]) j++;}i++;while(i < n && nums[i] == nums[i-1]) i++;}return res;}
};

本篇到此结束，感谢你的阅读。

祝你好运，向阳而生~