[leetcode]18.4Sum(Java实现)

leetcode地址：https://leetcode.com/problems/4sum/#/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.

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]]

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Java code：

package go.jacob.day619;import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class Demo1 {/* * accepted java O(n^3) solution based on 3sum * 基于3sum的解法 */public List<List<Integer>> fourSum(int[] nums, int target) {List<List<Integer>> result = new ArrayList<List<Integer>>();if (nums == null || nums.length < 4)return result;Arrays.sort(nums);for (int i = 0; i < nums.length - 3; i++) {/* * 后面两个if判断可以大大缩短测试时间：185ms到31ms */// first candidate too large, search finishedif (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target)break;// first candidate too smallif (nums[i] + nums[nums.length - 1] + nums[nums.length - 2] + nums[nums.length - 3] < target)continue;if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {for (int j = i + 1; j < nums.length - 2; j++) {// second candidate too largeif (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target)break; // second candidate too smallif (nums[i] + nums[j] + nums[nums.length - 1] + nums[nums.length - 2] < target)continue; if (j == i + 1 || (j > i + 1 && nums[j] != nums[j - 1])) {int sum = target - nums[i] - nums[j], left = j + 1, right = nums.length - 1;while (left < right) {if (nums[left] + nums[right] > sum)right--;else if (nums[left] + nums[right] < sum)left++;else {// 找到数对ArrayList<Integer> list = new ArrayList<Integer>();list.add(nums[i]);list.add(nums[j]);list.add(nums[left]);list.add(nums[right]);result.add(list);left++;right--;}while (left != j + 1 && left < right && nums[left] == nums[left - 1])left++;while (right != nums.length - 1 && right > left && nums[right] == nums[right + 1])right--;}}}}}return result;}}