《剑指offer》和为S的两个数字

本题来自《剑指offer》 和为S的两个数字

题目:

   输入一个递增排序的数组和一个数字S,在数组中查找两个数,使得他们的和正好是S,如果有多对数字的和等于S,输出两个数的乘积最小的。

思路:

   头尾双向夹逼遍历,题目是递增排序的数组,当和大于当前值是,那么就向前遍历,如果小于那么就向后遍历。

  要求是乘积最小,理论讲,两数相距最远,其乘积最小。

Python Code:

# -*- coding:utf-8 -*-
class Solution:
    def FindNumbersWithSum(self, array, tsum):
        # write code here
        result = []                                #result cache,empty
        if not isinstance(array, list):            #boundart condition judgement
            return result
        behind = 0                                 #first index of array
        ahead = len(array)-1                       #last index of array
        while behind<ahead:                        #the condition is behind small ahead
            cursum = array[behind]+array[ahead]    #summation
            if cursum==tsum:                       #if cursum equal target
                result.append(array[behind])       #then,append the value
                result.append(array[ahead])
                break
            elif cursum>tsum:                      #if cursum large target
                ahead -= 1                         #then,previous
            else:
                behind += 1                        #if cursum small target,then later
        return result
posted @ 2019-04-28 14:22  weilongyitian  阅读(172)  评论(0编辑  收藏  举报