LeetCode-15 3Sum(求3数和为零的情况总数)

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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:

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

思路：等同于求两个数和的相反数是否存在于这个数组中。可以延伸到K sum的问题。

public class Solution {    public List<List<Integer>> threeSum(int[] num) {    List<List<Integer>> list = new ArrayList<>();    Arrays.sort(num);    int start = 0;    while (start < num.length-2) {    if (start == 0 || num[start] != num[start-1] ) {    //留意判断条件，之前写的是start与start+1不等，导致一直不对。。    int instart = start+1;    int inlast = num.length-1;    while (instart < inlast) {if (num[instart] + num[inlast] == -num[start]) {//这里可以考虑用二分法查找来优化。List<Integer> temp = new ArrayList<Integer>();temp.add(num[start]);temp.add(num[instart]);temp.add(num[inlast]);list.add(temp);do {inlast--;//重复的数在这里直接跳过。} while (num[inlast] == num[inlast+1] && instart < inlast);do{instart++;} while(num[instart] == num[instart-1] && instart < inlast);}else if (num[instart] + num[inlast] > -num[start]) {inlast--;}else {instart++;}}    }    start++;}    return list;    }}

Runtime: 358 ms

更多收获可参考（个人觉得很棒）：Summary for LeetCode 2Sum, 3Sum, 4Sum, K Sum

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