3Sum

描述

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, $a \leq b \leq c$)
• 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)

分析

先排序，然后左右夹逼，复杂度 $O(n^2)$。

这个方法可以推广到k-sum，先排序，然后做k-2次循环，在最内层循环左右夹逼，时间复杂度是 $O(\max\{n \log n, n^{k-1}\})$。

代码

// 3Sum
// 先排序，然后左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)，Space Complexity: O(1)
public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums.length < 3) return result;
        Arrays.sort(nums);
        final int target = 0;

        for (int i = 0; i < nums.length - 2; ++i) {
            if (i > 0 && nums[i] == nums[i-1]) continue;
            int j = i+1;
            int k = nums.length-1;
            while (j < k) {
                if (nums[i] + nums[j] + nums[k] < target) {
                    ++j;
                    while(nums[j] == nums[j-1] && j < k) ++j;
                } else if(nums[i] + nums[j] + nums[k] > target) {
                    --k;
                    while(nums[k] == nums[k+1] && j < k) --k;
                } else {
                    result.add(Arrays.asList(nums[i], nums[j], nums[k]));
                    ++j;
                    --k;
                    while(nums[j] == nums[j-1] && j < k) ++j;
                    while(nums[k] == nums[k+1] && j < k) --k;
                }
            }
        }
        return result;
    }
};

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