[LeetCode]015. 3Sum-优快云博客

本文探讨了如何在整数数组中寻找三个数使它们的和为零的问题，并提出了一种有效的解决方案。采用一个for循环结合两数之和的方法（双指针），实现了O(n*n)的时间复杂度。特别注意去除重复解的处理方式。

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)

Solution:

use one 'for' loop, then use the two sum method(two pointers).

Running time: O(n*n);

Notice: must eliminate the duplicates of num[start], num[end](in the while loop), and the num[i](out of the while loop, in the for loop);

代码的39和40行需要注意，之前是写成这样：

if(i<num.length-1){
    while(num[i] == num[i+1]){
        i++;
    }
}

结果[0,0,0]的test case就直接无法通过，用eclipse调试才发现是这里的out of bounday错误。

感觉if 里面写while是大忌啊，一般都是while里面写if，不过这里while里写if也是会出错，边界不容易控制。需要注意。

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        // Start typing your Java solution below
        // DO NOT write main() function
        Arrays.sort(num);
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length < 3){
            return list;
        }
        if(num[num.length-1] < 0){
            return list;
        }
        for(int i=0; i<num.length; i++){
            int target = - num[i];
            int start = i+1;
            int end = num.length - 1;
            while(start < end){
                if(num[start] + num[end] < target){
                    start++;
                }else if(num[start] + num[end] > target){
                    end--;
                }else{
                    ArrayList<Integer> al = new ArrayList<Integer>();
                    al.add(num[i]);
                    al.add(num[start]);
                    al.add(num[end]);
                    list.add(al);
                    start++;
                    end--;
                    while(start<end && num[start] == num[start-1]){
                        start++;
                    }
                    while(start<end && num[end] == num[end+1]){
                        end--;
                    }
                    // remove the duplicates of start and end;
                }
            }
            while(i<num.length-1 && num[i] == num[i+1]){
                i++;
            }
            //remove the duplicates of target;
        }
        return list;
    }
}