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+# [18. 4Sum](https://leetcode.com/problems/4sum/)
+# 思路
+和[3sum](https://github.com/ShusenTang/LeetCode/blob/master/15.%203Sum.md)这题基本一样的，注意这题由于循环层数比较多所以如果能
+在外层循环判断一下的话可以提前终止循环或跳过某次循环，见代码。   
+时间复杂度O(n^3)
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<vector<int>> fourSum(vector<int>& nums, int target) {
+        vector<vector<int>>res;
+        int len = nums.size(), low, high, sum;
+        if(len < 4) return res;
+        sort(nums.begin(), nums.end());
+        for(int i = 0; i < len - 3; i++){
+            if(i != 0 && nums[i] == nums[i - 1]) continue;
+            
+            // 提前退出或跳过不可能的情况
+            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) break; // 最小的都比target大了，可以提前终止循环
+            if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1] < target) continue;
+            
+            for(int j = i + 1; j < len - 2; j++){
+                if(j != i + 1 && nums[j] == nums[j - 1]) continue;
+                low = j + 1;
+                high = len - 1;
+                while(low < high){
+                    sum = nums[i] + nums[j] + nums[low] + nums[high];
+                    if(sum < target) 
+                        while(++low < high && nums[low] == nums[low - 1]) ; // 不断右移low指针
+                    else if(sum > target) 
+                        while(low < --high && nums[high] == nums[high + 1]) ; // 不断左移high指针
+                    else{
+                        vector<int>tmp = {nums[i], nums[j], nums[low++], nums[high--]};
+                        res.push_back(tmp);
+                        while(low < high && nums[low] == nums[low - 1]) low++;
+                        while(low < high && nums[high] == nums[high + 1]) high--;
+                    }
+                }
+            }
+        }
+        return res;
+        
+    }
+};
+```