[LeetCode] 15. 3Sum 三数之和

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

解法：排序 + 双指针。先对数组排序，然后遍历排序后的数组, 当循环到nums[i]时，对后面的数使用双指针left和right分别指向第一个和最后一个数，如果3个数的和等于0，就找到一组添加到结果中；如果小于0，说明和小了，要往大的方向移动，left指针右移1位；如果和大于0，说明大了，right指针左移1位。再看3个数的和。由于题目要求不能有重复的答案，所以对于每个指针移动后都要看是否和前面的相等，如果相等则跳过。

Time: O(n^2)  Space: O(1)

Java:

public List<List<Integer>> threeSum(int[] num) {
    Arrays.sort(num);
    List<List<Integer>> res = new LinkedList<>(); 
    for (int i = 0; i < num.length-2; i++) {
        if (i == 0 || (i > 0 && num[i] != num[i-1])) {
            int lo = i+1, hi = num.length-1, sum = 0 - num[i];
            while (lo < hi) {
                if (num[lo] + num[hi] == sum) {
                    res.add(Arrays.asList(num[i], num[lo], num[hi]));
                    while (lo < hi && num[lo] == num[lo+1]) lo++;
                    while (lo < hi && num[hi] == num[hi-1]) hi--;
                    lo++; hi--;
                } else if (num[lo] + num[hi] < sum) lo++;
                else hi--;
           }
        }
    }
    return res;
}

Python:

def threeSum(self, nums):
    res = []
    nums.sort()
    for i in xrange(len(nums)-2):
        if i > 0 and nums[i] == nums[i-1]:
            continue
        l, r = i+1, len(nums)-1
        while l < r:
            s = nums[i] + nums[l] + nums[r]
            if s < 0:
                l +=1 
            elif s > 0:
                r -= 1
            else:
                res.append((nums[i], nums[l], nums[r]))
                while l < r and nums[l] == nums[l+1]:
                    l += 1
                while l < r and nums[r] == nums[r-1]:
                    r -= 1
                l += 1; r -= 1
    return res　

Python:

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums, result, i = sorted(nums), [], 0
        while i < len(nums) - 2:
            if i == 0 or nums[i] != nums[i - 1]:
                j, k = i + 1, len(nums) - 1
                while j < k:
                    if nums[i] + nums[j] + nums[k] < 0:
                        j += 1
                    elif nums[i] + nums[j] + nums[k] > 0:
                        k -= 1
                    else:
                        result.append([nums[i], nums[j], nums[k]])
                        j, k = j + 1, k - 1
                        while j < k and nums[j] == nums[j - 1]:
                            j += 1
                        while j < k and nums[k] == nums[k + 1]:
                            k -= 1
            i += 1
        return result　　

Python: wo

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        nums = sorted(nums)
        n = len(nums)
        i = 0
        while i < n - 2:
            j, k = i + 1, n - 1
            while j < k:                
                s = nums[i] + nums[j] + nums[k]
                if s > 0:
                    k -= 1
                elif s < 0:
                    j += 1
                else:
                    res.append([nums[i], nums[j], nums[k]])
                    j += 1
                    k -= 1
                    while nums[j-1] == nums[j] and j < k:
                        j += 1
                    while nums[k+1] == nums[k] and j < k :
                        k -= 1                                
            i += 1
            while nums[i-1] == nums[i] and i < n - 2:
                i += 1
            
        return res　　

C++：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        for (int k = 0; k < nums.size(); ++k) {
            if (nums[k] > 0) break;
            if (k > 0 && nums[k] == nums[k - 1]) continue;
            int target = 0 - nums[k];
            int i = k + 1, j = nums.size() - 1;
            while (i < j) {
                if (nums[i] + nums[j] == target) {
                    res.push_back({nums[k], nums[i], nums[j]});
                    while (i < j && nums[i] == nums[i + 1]) ++i;
                    while (i < j && nums[j] == nums[j - 1]) --j;
                    ++i; --j;
                } else if (nums[i] + nums[j] < target) ++i;
                else --j;
            }
        }
        return res;
    }
};

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[LeetCode] 16. 3Sum Closest 最近三数之和

[LeetCode] 259. 3Sum Smaller 三数之和较小值

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