标签：nbsp obj 移动 部分 nice pretty style target ack
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: 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] ]
易错：出现重复的结果
暴力枚举三个数复杂度为O(N^3)。先考虑2Sum的做法，假设升序数列a，对于一组解ai,aj, 另一组解ak,al ，必然满足 i<k j>l 或 i>k j<l, 因此我们可以用两个指针，初始时指向数列两端
指向数之和大于目标值时，右指针向左移使得总和减小，反之左指针向右移。由此可以用O(N)的复杂度解决2Sum问题，3Sum则枚举第一个数O(N^2)，使用有序数列的好处是，在枚举和移动指针时值相等的数可以跳过，省去去重部分
Python实现：
class Solution(object): def threeSum(self, nums): nums.sort() ans = [] # 结果 for i in range(0, len(nums)-2): if i and nums[i] == nums[i-1]: continue target = -1*nums[i] left = i + 1 right = len(nums)-1 while left < right: if nums[left] + nums[right] == target: ans.append([nums[i], nums[left], nums[right]]) left += 1 right -= 1 while left < right and nums[left] == nums[left-1]: #避免结果重复 ！！！ left += 1 while left < right and nums[right] == nums[right+1]: right -= 1 elif nums[left] + nums[right] < target: left += 1 # while left < right and nums[left] == nums[left+1]: # left += 1 else: right -= 1 # while left < right and nums[right] == nums[right-1]: # right -= 1 return ans
标签：nbsp obj 移动 部分 nice pretty style target ack
原文地址：http://www.cnblogs.com/ytq1016/p/6863520.html
