[LeetCode] 18. 4Sum

传送门

Description

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.

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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] ]

 

思路

题意：给出一串整数值序列，输出a + b + c + d = target的方案

题解：此题与3Sum类似，区别在于这是求取四个数的和等于目标值的方案，因此将其降维为求取3Sum问题。

+ View Code?

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class Solution {     //61ms     public List<List<Integer>> fourSum(int[] nums, int target) {         Arrays.sort(nums);         List<List<Integer>>res = new ArrayList<>();         int len = nums.length;         for (int i = 0;i < len - 3;i++){             threeSum(nums,nums[i],target - nums[i],i + 1,len,res);             while (i + 1 < len && nums[i + 1] == nums[i]){                 i++;             }         }         return res;     }       public void threeSum(int[] nums,int val,int target,int lo,int ro,List<List<Integer>>res) {           for (int i = lo;i < ro - 2;i++){             int left = i + 1,right = ro - 1;             int sum = target - nums[i];             while (left < right){                 if (nums[left] + nums[right] < sum){                     left++;                 } else if (nums[left] + nums[right] > sum){                     right--;                 } else{                     res.add(Arrays.asList(val,nums[i],nums[left],nums[right]));                     while (++left < right && nums[left] == nums[left - 1]){}                     while (--right > left && nums[right] == nums[right + 1]){}                 }             }             while (i + 1 < ro - 2 && nums[i + 1] == nums[i]){                 i++;             }         }       }   }