leetcode016 017

给一组数求 相加的0的n组数
先排序
然后确定第一个数
第二个数和第三个数从两边往中间挪
如果三数相加大于0
因为第二个数已经是最小的了
所以把第三个数往左挪
同理 小于0 第二个数往右挪
这样复杂度是O（n方）

class Solution:    # @param {integer[]} nums    # @return {integer[][]}    def threeSum(self,nums):        nums.sort()        l = len(nums)        res = []        if l < 2:            return res        i = 0        while i < l - 2:            j = i + 1            k = l - 1            while j < k:                ##print i , j , k                if nums[i]+nums[k]+nums[j]>0:                    k = k -1                elif nums[i]+nums[k]+nums[j]<0:                    j = j + 1                else:                    res.append([nums[i],nums[j],nums[k]])                    while j < k and nums[j] == nums[j+1]:                        j = j + 1                    while j < k and nums[k] == nums[k-1]:                        k = k -1                    j = j + 1                    k = k - 1            while i < l - 2 and nums[i] == nums[i+1]:                i = i + 1            i = i + 1        return res 

17题

求一组数中三个数相加的离目标数最近的那个和
比第一题简单 同理

def threeSum(nums , target):    nums.sort()    l = len(nums)    res = nums[0]+nums[1]+nums[2]    if l < 2:        return res    i = 0    while i < l - 2:        j = i + 1        k = l - 1        while j < k:            num =nums[i]+nums[j]+nums[k]            if abs(res-target)>abs(num-target):                res = num            if nums[i]+nums[k]+nums[j]>target:                k = k -1            elif nums[i]+nums[k]+nums[j]<target:                j = j + 1            else:                while j < k and nums[j] == nums[j+1]:                    j = j + 1                while j < k and nums[k] == nums[k-1]:                    k = k -1                j = j + 1                k = k - 1        while i< l - 2 and nums[i] == nums[i+1]:            i = i + 1        i = i + 1    return res###print threeSum([-1,2,1,-4],1)

贵在坚持