1. 移动零

void moveZeroes(vector<int>& nums)
{size_t  front = 0;size_t back = 0;while(back < nums.size() && nums[back] != 0){++back;}front = back++;while(back < nums.size()){if(0 == nums[back]){++back;}else{swap(nums[front++], nums[back++]);     }}
}

2. 复写零
   先找到最后一个被“复写”的数，然后从后向前完成“复写”操作。同时要注意处理一下边界情况。

void duplicateZeros(vector<int>& arr)
{int front = 0;int back = -1;while(true){if(0 == arr[front]){back += 2;}else{back += 1;}if(back == arr.size()){--back;arr[back--] = 0;--front;break;}if(back == arr.size() - 1){break;}++front;}while(front >= 0){if(arr[front] == 0){arr[back--] = 0;arr[back--] = 0;--front;}else{arr[back--] = arr[front--];}}
}

3. 快乐数

// 方法一
bool isHappy(int n)
{unordered_set<int> res({n});int num = 0;while(true){while(n != 0){num += pow((n % 10), 2);n /= 10;}if(num == 1){return true;}else if(res.find(num) != res.end()){return false;}else{res.insert(num);}n = num;num = 0;}
}
// 方法二
int Run(int num)
{int ret = 0;while(num != 0){ret += pow((num % 10), 2);num /= 10;}return ret;
}bool isHappy(int n)
{int slow = n;int fast = n;while (true){slow = Run(slow);fast = Run(fast);fast = Run(fast);if (slow == fast && fast == 1){return true;}else if (slow == fast){return false;}}
}

4. 盛最多水的容器

int maxArea(vector<int>& height)
{size_t left= 0;size_t right = height.size() - 1;size_t ret = (right - left) * min(height[left], height[right]);size_t tmp = 0;size_t area = 0;do{if(height[left] < height[right]){tmp = height[left];while(left < right && height[++left] <= tmp);}else{tmp = height[right];while(left < right && height[--right] <= tmp);}if(left < right){area = (right - left) * min(height[left], height[right]);if(area > ret) ret = area;}}while(left < right);return ret;
}

5. 有效三角形的个数
   利用单调性，先固定最大的数，在最大的数的左区间内使用双指针算法。

inline bool check(int side1, int side2, int side3)
{return (side1 + side2 > side3);
}
int triangleNumber(vector<int>& nums)
{if(nums.size() < 3) return 0;sort(nums.begin(), nums.end());size_t max_index = nums.size() - 1;size_t left_index = 0;size_t right_index = max_index - 1;int ret = 0;while(max_index > 1){while(left_index < right_index){if(check(nums[left_index], nums[right_index], nums[max_index])){ret += (right_index - left_index);--right_index;}else{++left_index;}}--max_index;left_index = 0;right_index = max_index - 1;}return ret;
}

6. 查找总价格为目标值的两个商品

vector<int> twoSum(vector<int>& price, int target)
{size_t left = 0;size_t right = price.size() - 1;vector<int> v(2);while(left < right){if(price[left] + price[right] < target){++left;}else if(price[left] + price[right] > target){--right;}else{v[0] = price[left];v[1] = price[right];break;}}return v;
}

7. 三数之和
   注意对去重的处理（跳过重复元素）。

vector<vector<int>> threeSum(vector<int>& nums)
{vector<vector<int>> vv;sort(nums.begin(), nums.end());size_t max_index = nums.size() - 1;size_t left_index = 0;size_t right_index = max_index - 1;while(max_index > 1){while(left_index < right_index){if(nums[left_index] + nums[right_index] + nums[max_index] > 0){while(left_index < --right_index && nums[right_index] == nums[right_index + 1]);}else if(nums[left_index] + nums[right_index] + nums[max_index] < 0){while(++left_index < right_index && nums[left_index] == nums[left_index - 1]);}else{vv.push_back({nums[left_index], nums[right_index], nums[max_index]});while(left_index < --right_index && nums[right_index] == nums[right_index + 1]); // or ++left_index}}while(--max_index > 1 && nums[max_index] == nums[max_index + 1]);left_index = 0;right_index = max_index - 1;}return vv;
}

8. 四数之和
   结合三数之和解题。

long long Sum(long long num1, long long num2, long long num3)
{return num1 + num2 + num3;
}
void threeSum(vector<vector<int>>& vv, vector<int>& nums, size_t max_index, int max, int target)
{size_t left_index = 0;size_t right_index = max_index - 1;while(max_index > 1){while(left_index < right_index){if(Sum(nums[left_index], nums[right_index], nums[max_index]) > target){while(left_index < --right_index && nums[right_index] == nums[right_index + 1]);}else if(Sum(nums[left_index], nums[right_index], nums[max_index]) < target){while(++left_index < right_index && nums[left_index] == nums[left_index - 1]);}else{vv.push_back({nums[left_index], nums[right_index], nums[max_index], max});while(left_index < --right_index && nums[right_index] == nums[right_index + 1]);}}while(--max_index > 1 && nums[max_index] == nums[max_index + 1]);left_index = 0;right_index = max_index - 1;}
}
vector<vector<int>> fourSum(vector<int>& nums, int target)
{   vector<vector<int>> vv;if(nums.size() < 4) return vv;sort(nums.begin(), nums.end()); // 排序size_t max_index = nums.size() - 1; // 固定第4个数while(max_index > 2){threeSum(vv, nums, max_index - 1, nums[max_index], target - nums[max_index]);while(--max_index > 2 && nums[max_index] == nums[max_index + 1]);}return vv;
}