[LeetCode] 15. 3Sum ☆☆☆(3数和为0)

描述

Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

解析

先排序，再确定一个数字，用后面的数据获取0 - num[i]的值，转成 2sum的问题。要注意跳过重复数据。

代码

解法1(解析所述)

public static List<List<Integer>> threeSum3(int[] num) {
        if (num == null || num.length < 3) {
            return new ArrayList<>();
        }
        Arrays.sort(num);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < num.length - 2; i++) {
            if (i == 0 || (i > 0 && num[i] != num[i - 1])) {
                int target = 0 - num[i];
                int start = i + 1;
                int end = num.length - 1;
                while (start < end) {
                    if (num[start] + num[end] == target) {
                        res.add(Arrays.asList(num[i], num[start], num[end]));
                        //去除重复元素
                        while (start < end && num[start] == num[start + 1]) {
                            start++;
                        }
                        while (start < end && num[end] == num[end - 1]) {
                            end--;
                        }
                        start++;
                        end--;
                    } else if (num[start] + num[end] > target) {
                        end--;
                    } else {
                        start++;
                    }
                }
            }
        }
        return res;
    }

数组中找等于指定数的集合

也可以用数组num，在其中找n个数等于指定数的解法。