【LeetCode】18. 4Sum

18. 4Sum

题目描述

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]]

解题思路

双指针问题。
思路还是与3Sum问题相似。差异在于，此处是依次固定第一和第二个数字，然后通过双指针移动控制第三和第四个数字；而3Sum则只需固定第一个数字。
注意：要仔细分析所有while循环的控制条件，防止出现死循环。

AC代码

class Solution {public:    vector<vector<int> > fourSum(vector<int>& nums, int target) {        sort(nums.begin(), nums.end());        vector<vector<int> > ans;        const int vSize = nums.size();        //fix the first number        for (int first = 0; first < vSize - 3; ++first) {            //fix the second number            for (int second = first + 1; second < vSize - 2; ++second) {                int third = second + 1;                int forth = vSize - 1;                while (third < forth) {                    int curSum = nums[first] + nums[second] + nums[third] + nums[forth];                    if (curSum < target)                        third++;                    else if (curSum > target)                        forth--;                    else {                        vector<int> tempAns;                        //注意此处push_back的是数值nums[first],而非下标first                        tempAns.push_back(nums[first]);                        tempAns.push_back(nums[second]);                        tempAns.push_back(nums[third]);                        tempAns.push_back(nums[forth]);                        ans.push_back(tempAns);                        //此处不可以直接写nums[third+1]==nums[third]                        //因为要确保third一定会被更新到                        while (third < forth && nums[third] == tempAns[2]) third++;                        while (third < forth && nums[forth] == tempAns[3]) forth--;                    }                }                while (second < vSize - 1 && nums[second + 1] == nums[second]) second++;            }            while (first < vSize - 1 && nums[first + 1] == nums[first]) first++;        }        return ans;    }};