LeetCode-3Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/chilakaka/article/details/44030301

本文介绍LeetCode经典题目3Sum的解决方案，通过双指针技巧实现O(n^2)的时间复杂度，同时确保输出不含重复三元组。文章详细解析了如何遍历数组并调整指针位置来寻找所有可能的零和三元组。

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https://oj.leetcode.com/problems/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:

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)

典型的双指针问题，把0-n[i]作为target，然后在剩余的数组中做2Sum就可以了。时间复杂度O(n^2).

因为输出中不能用重复，每次start++, end--都要把重复数字略过，外循环中i++的时候也要把重复数字略过。

public class Solution {
    public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> rst = new ArrayList<List<Integer>>();
        if(num==null || num.length<3) return rst;
        Arrays.sort(num);
        for(int i=0; i<num.length; i++){
            int start = i+1;
            int end = num.length-1;
            int target = -num[i];
            while(start<end){
                int sum= num[start]+num[end];
                if(sum<target){
                    start++;
                    while(start<num.length && num[start]==num[start-1]) start++;
                } 
                else if(sum>target){
                    end--;
                    while(end>=0 && num[end]==num[end+1]) end--;
                } 
                else{
                    List<Integer> sol = new ArrayList<Integer>();
                    sol.add(num[i]);
                    sol.add(num[start]);
                    sol.add(num[end]);
                    rst.add(sol);
                    start++;
                    while(start<num.length && num[start]==num[start-1]) start++;
                    end--;
                    while(end>=0 && num[end]==num[end+1]) end--;
                }
            }
            while(i<(num.length-1) && num[i+1]==num[i]) i++;
        }
        return rst;
    }
}