15. 3Sum

来源：互联网发布：lable java 编辑：程序博客网时间：2024/06/08 09:33

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: 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：

首先将输入的数字全部进行一次sort()排序，而后利用两个指针，i,j，i是从num[0]开始，j是从num[n-1]开始，依次对比，找到a+b+c=0的-c为target。

class Solution:    global res    res=[]    def Sum3(self, num):        for i in range(0, num.__len__()):            if i > 0 and num[i] == num[i - 1]:                pass            sum2(num, res, i)        return res    def sum2(self, num, res, target):        i = target + 1;        j = num.__len__ - 1        while i < j:            if num[target] + num[i] + num[j] < 0:                i += 1            elif num[target] + num[i] + num[j] > 0:                j -= 1            else:                a = []                a.append(num[target])                a.append(num[i])                a.append(num[j])                res = a                i += 1                j -= 1                while i < num.__len__ and num[i] == num[i - 1]:                    i += 1                while j >= 0 and num[j] == num[j + 1]:                    j -= 1

num=[]
n=int(input("请输入数组的长度："))
请输入数组的长度：>? 5
for i in range(0, n):
    temp = int(input("输入数字："))
    num.append(temp)
    num.sort()

sl=Solution()

sl.Sum3(num)

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