leetcode 15. 3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

在一组数组中找出所有的不重复的 [a,b,c],使a+b+c=0 (a <= b <= c).

先排序,然后遍历数组,固定一个数为a,其后一个数为b,最后一个数为c,若符合题意则保存;若小于零则说明b太小,将b后的数作为b,再比较;若大于零则说明c太大,将c前面的数作为c,再比较.

public class A15_3Sum {public List<List<Integer>> threeSum(int[] nums) {Arrays.sort(nums);List<List<Integer>> ans = new ArrayList<List<Integer>>();for(int i = 0; i < nums.length - 2; i++) {if(nums[i] > 0) break;int second = i + 1;int third = nums.length - 1;while(second < third) {List<Integer> list3 = new ArrayList<Integer>();int sum = nums[i] + nums[second] + nums[third];if(sum == 0) {list3.add(nums[i]);list3.add(nums[second]);list3.add(nums[third]);ans.add(list3);do {second++;} while(second < third && nums[second - 1] == nums[second]);do {third--;} while(second < third && nums[third + 1] == nums[third]);} else if(sum > 0) {do {third--;} while(second < third && nums[third + 1] == nums[third]);} else {do {second++;} while(second < third && nums[second - 1] == nums[second]);}}while(i < nums.length - 2 && nums[i] == nums[i + 1])i++;}return ans;    }}