15. 3Sum

题目：

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)

链接： http://leetcode.com/problems/3sum/

题解：

3 pointers，注意处理重复的情况。

Time Complexity - O(n²), Space Complexity - O(1)

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num.length < 3)
            return result;
        Arrays.sort(num);
        
        for(int start = 0; start < num.length - 2; start ++){
            if(start > 0 && num[start] == num[start - 1])
                continue;
            int left = start + 1, right = num.length - 1;
            
            while(left < right){
                int sum = num[start] + num[left] + num[right];
                if(sum == 0){
                    ArrayList<Integer> list = new ArrayList<Integer>();
                    list.add(num[start]);
                    list.add(num[left]);
                    list.add(num[right]);
                    result.add(list);
                    left ++;
                    right --;
                    while(left < right && num[left] == num[left - 1])
                        left ++;
                    while(left < right && num[right] == num[right + 1])
                        right --;
                } else if (sum < 0){
                    left ++;
                } else {
                    right --;
                }
            }
        }
        
        return result;
    }
}

 

二刷：

为满足题意，先对数组进行排序。注意去重，去重的部分包括一个主指针i， 两个分指针lo以及hi，都要考虑。在加结果的时候，也可以使用Arrays.asList(num[i], num[lo], num[hi])，但之后这个list就是定长的，不可以再被增加了

Java:

Time Complexity - O(n²)，  Space Complexity - O(n)

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length < 3) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }    
            int lo = i + 1;
            int hi = nums.length - 1;
            while (lo < hi) {
                int sum = nums[i] + nums[lo] + nums[hi];
                if (sum == 0) {
                    List<Integer> list = new ArrayList<>();
                    list.add(nums[i]);
                    list.add(nums[lo]);
                    list.add(nums[hi]);
                    res.add(list);
                    lo++;
                    hi--;
                    while (lo < hi && nums[lo] == nums[lo - 1]) {
                        lo++;
                    }
                    while (lo < hi && nums[hi] == nums[hi + 1]) {
                        hi--;
                    }
                } else if (sum < 0) {
                    lo++;
                } else {
                    hi--;
                }
            }
        }
        
        return res;
    }
}

Python:

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        if not nums:
            return res
        nums.sort();
        for i in range(0, len(nums) - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            lo = i + 1
            hi = len(nums) - 1
            while lo < hi:
                sum = nums[i] + nums[lo] + nums[hi]
                if sum == 0:
                    res.append([nums[i], nums[lo], nums[hi]])
                    lo += 1
                    hi -= 1
                    while lo < hi and nums[lo] == nums[lo - 1]:
                        lo += 1
                    while lo < hi and nums[hi] == nums[hi + 1]:
                        hi -= 1
                elif sum < 0:
                    lo += 1
                else:
                    hi -= 1
        return res
                    

 

三刷:

有两种思路。

1. 先排序再用双指针

2. 使用Two Sum的方法， 遍历 + HashMap

Java:

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null) return res;
        Arrays.sort(nums);
        
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int lo = i + 1;
            int hi = nums.length - 1;
            while (lo < hi) {
                int sum = nums[i] + nums[lo] + nums[hi];
                if (sum == 0) {
                    List<Integer> list = new ArrayList<>();
                    list.add(nums[i]);
                    list.add(nums[lo]);
                    list.add(nums[hi]);
                    res.add(list);
                    lo++;
                    hi--;
                    while (lo < hi && nums[lo] == nums[lo - 1]) lo++;
                    while (lo < hi && nums[hi] == nums[hi + 1]) hi--;
                } else if (sum < 0) {
                    lo++;
                } else {
                    hi--;
                }
            }
        }
        return res;
    }
}

 

 

Reference:

https://leetcode.com/discuss/67914/easiest-java-solution

https://leetcode.com/discuss/23638/concise-o-n-2-java-solution