15. 三数之和

算法-目录
题目来源：Leetcode 15. 三数之和

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """      
        n=len(nums)
        res=[]
        if(not nums or n<3):
            return []
        nums.sort()
        res=[]
        for i in range(n):
            if(nums[i]>0):
                return res
            if(i>0 and nums[i]==nums[i-1]):
                continue
            L=i+1
            R=n-1
            while(L<R):
                if(nums[i]+nums[L]+nums[R]==0):
                    res.append([nums[i],nums[L],nums[R]])
                    while(L<R and nums[L]==nums[L+1]):
                        L=L+1
                    while(L<R and nums[R]==nums[R-1]):
                        R=R-1
                    L=L+1
                    R=R-1
                elif(nums[i]+nums[L]+nums[R]>0):
                    R=R-1
                else:
                    L=L+1
        return res

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        n = len(nums)
        nums.sort()
        ans = list()
        for a in range(n):
            if a>0 and nums[a]==nums[a-1]:
                continue
            target=-nums[a]
            c=n-1
            for b in range(a+1,n):
                if b>a+1 and nums[b]==nums[b-1]:
                    continue
                while b<c and nums[b]+nums[c]>target:
                    c-=1
                if b==c:
                    break
                if nums[b]+nums[c]==target:
                    ans.append([nums[a],nums[b],nums[c]])
        return ans

# 注解版
class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        n = len(nums)
        nums.sort()
        ans = list()
        
        # 枚举 a
        for first in range(n):
            # 需要和上一次枚举的数不相同
            if first > 0 and nums[first] == nums[first - 1]:
                continue
            # c 对应的指针初始指向数组的最右端
            third = n - 1
            target = -nums[first]
            # 枚举 b
            for second in range(first + 1, n):
                # 需要和上一次枚举的数不相同
                if second > first + 1 and nums[second] == nums[second - 1]:
                    continue
                # 需要保证 b 的指针在 c 的指针的左侧
                while second < third and nums[second] + nums[third] > target:
                    third -= 1
                # 如果指针重合，随着 b 后续的增加
                # 就不会有满足 a+b+c=0 并且 b
                if second == third:
                    break
                if nums[second] + nums[third] == target:
                    ans.append([nums[first], nums[second], nums[third]])
        
        return ans