leetcode015 3Sum

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题目

15. 3 Sum
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)

思路：

我用了一种最笨的方法：用三个循环硬解。当然如果什么都不处理，肯定是会TLE的。我先把数组用Arrays的sort方法排序，然后再循环的过程中如果前后元素相同则跳过。

代码：

public List<List<Integer>> threeSum(int[] nums){    Arrays.sort(nums);    List<List<Integer>> lists = new ArrayList<>();    for(int i = 0; i < nums.length; i++)    {        for(int j = i + 1; j < nums.length; j++)        {            for(int k = j + 1; k < nums.length; k++)            {                if(nums[i] + nums[j] + nums[k] == 0)                {                    lists.add(Arrays.asList(nums[i], nums[j], nums[k]));                }                while(k+1 < nums.length&& nums[k] == nums[k+1])                    k++;            }            while(j+1 < nums.length&& nums[j] == nums[j+1])                j++;        }        while(i+1 < nums.length&& nums[i] == nums[i+1])            i++;    }    return lists;}

结果细节（图）：

更新：

代码

public List<List<Integer>> threeSum(int[] nums){    Arrays.sort(nums);    List<List<Integer>> lists = new ArrayList<>();    for(int i = 0; i < nums.length-2; i++)    {        if(i == 0 || nums[i] != nums[i-1])        {            int low = i+1;            int high = nums.length-1;            int sum = -nums[i];            while(low < high)            {                if(nums[low]+ nums[high] == sum)                {                    lists.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 lists;}

细节图

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