18. 4Sum

Given an array S of n integers, are there elements a,b,c, andd inS such that a + b +c +d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,a ≤b ≤c ≤d)
• 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)

public class Solution {public List<List<Integer>> fourSum(int[] nums, int target) {List<List<Integer>> result = new ArrayList<List<Integer>>();Arrays.sort(nums);kSum(result, new ArrayList<Integer>(), nums, target, 4, 0, nums.length-1);return result;}private void kSum(List<List<Integer>> result,List<Integer> cur, int[] nums, int target,int k, int start, int end) {if(k == 0 || nums.length ==0 || start>end)return;if(k == 1){for (int i = start; i <= end; i++) {if (nums[i] == target) {cur.add(nums[i]);//添加result.add(new ArrayList<Integer>(cur));//放入到结果中cur.remove(cur.size()-1);//移除}}return;}if(k == 2){int sum;while (start < end){sum = nums[start] + nums[end];if(sum == target){cur.add(nums[start]);cur.add(nums[end]);result.add(new ArrayList<Integer>(cur));cur.remove(cur.size()-1);cur.remove(cur.size()-1);//避免重复while(start < end && nums[start] == nums[start+1])++start;++start;while(start < end && nums[end] == nums[end-1])--end;--end;}else if(sum < target){//避免重复while(start < end && nums[start] == nums[start+1])++start;++start;}else{while(start < end && nums[end] == nums[end-1])--end;--end;}}return;}//每次务必会进行数据预处理，简单的情况判断，避免了不必要的情况if(k*nums[start] > target || k*nums[end] < target) return;for(int i = start; i <= (end-k+1); i++){if(i > start && nums[i] == nums[i-1])continue;//避免了重复if(k*nums[i] <= target) {cur.add(nums[i]);kSum(result, cur, nums, target - nums[i], k - 1, i + 1, end);//开始了神奇的递归了cur.remove(cur.size() - 1);//输入的地方总要对应一个输出}}}}//k=0,1太简单了,k=2,见上一题

这题跟之前的几道题是差不多的~