【剑指offer】和为S的两个数

题目描述

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：

方法一：粗暴遍历法

# -*- coding:utf-8 -*-
class Solution:
    def FindNumbersWithSum(self, array, tsum):
        # write code here
        if not array or not tsum:
            return array
        for i in range(len(array)):
            res = tsum - array[i]
            if res in array[0:i]+array[i+1:]:
                return [array[i],res]
        return []

方法二：数组前后两个指针

# -*- coding:utf-8 -*-
class Solution:
    def FindNumbersWithSum(self, array, tsum):
        # write code here
        if not array or not tsum:
            return array
        small = 0
        big = len(array) - 1
        minn = 100000
        res = []
        cur_sum = array[small] + array[big]
        while array[small] < tsum/2:
            if cur_sum == tsum:
                if minn > array[small]*array[big]:
                    res = [array[small],array[big]]
                    minn = array[small]*array[big]
            while cur_sum < tsum and array[small] < tsum/2:
                    small += 1
                    cur_sum = array[small] + array[big]
                    if cur_sum == tsum:
                        if minn > array[small]*array[big]:
                            res = [array[small],array[big]]
                            minn = array[small]*array[big]
            big -= 1
            cur_sum = array[small] + array[big]
        return res