【Leetcode】4Sum (Sum)

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:

• 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)

这道题就是将4sum转化为3sum再转化为2sum进行求解，理解透转化的过程是这道题求解最重要的部分

直接上代码 来自Code_Ganker

public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();if (num == null || num.length < 4)return result;Arrays.sort(num);for (int i = num.length - 1; i > 2; i--) {if (i == num.length - 1 || num[i] != num[i + 1]) {ArrayList<ArrayList<Integer>> curRes = threeSum(num, i - 1,target - num[i]);for (int j = 0; j < curRes.size(); j++) {curRes.get(j).add(num[i]);}result.addAll(curRes);}}return result;}public ArrayList<ArrayList<Integer>> threeSum(int[] num, int end, int target) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();for (int i = end; i > 1; i--) {if (i == end || num[i] != num[i + 1]) {ArrayList<ArrayList<Integer>> curRes = twoSum(num, i - 1,target - num[i]);for (int j = 0; j < curRes.size(); j++) {curRes.get(j).add(num[i]);}result.addAll(curRes);}}return result;}public ArrayList<ArrayList<Integer>> twoSum(int[] num, int end, int target) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();int left = 0;int right = end;while (left < right) {if (num[left] + num[right] == target) {ArrayList<Integer> temp = new ArrayList<Integer>();temp.add(num[left]);temp.add(num[right]);result.add(temp);left++;right--;while (left < right && num[left] == num[left - 1])left++;while (left < right && num[right] == num[right + 1])right--;} else if (num[left] + num[right] > target)right--;elseleft++;}return result;}