15. 3Sum

题目分析

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]
]

解这道题目，首先我们将数组排序，然后从头开始遍历数组，将 3Sum 问题转成 Two Sum 问题。
时间复杂度为O(n ^ 2)

代码

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        // 只需要遍历到倒数第三个元素
        Arrays.sort(nums);
        for(int i = 0; i < nums.length - 2; i++) {
            if(i > 0 && nums[i] == nums[i - 1]) continue;
            // 问题转成了在 low 到 high 区间的 two sum 问题
            int low = i + 1, high = nums.length - 1, sum = 0 - nums[i];
            while(low < high) {
                if(nums[low] + nums[high] == sum) {
                    res.add(Arrays.asList(nums[i], nums[low], nums[high]));
                    while(low < high && nums[low] == nums[low + 1]) low ++;
                    while(low < high && nums[high] == nums[high - 1]) high --;
                    low ++;
                    high --;
                } else if(nums[low] + nums[high] < sum) {
                    low ++;
                } else {
                    high --;
                }
            }
        }
        return res;
    }
}