[coding horror] 1 - sum 2

sum 2

描述

輸入一個遞增排序的數組和一個數字S，在數組中查找兩個數，是的他們的和正好是S，若是有多對數字的和等於S，輸出兩個數的乘積最小的。

對應每一個測試案例，輸出兩個數，小的先輸出。

coding horror

class Solution {
public:
    vector<int> FindNumbersWithSum(vector<int> array,int sum) {
        vector<int> ret;

        if(array.size()==0){
            return ret;
        };

        vector<int> first_nums;
        vector<int> second_nums;

        int first_index = 0;
        int second_index = array.size() - 1;

        while(first_index < second_index){ // index越界
            int first = array[first_index];
            int second = array[second_index];

            if( first + second == sum) {
                first_nums.push_back(first);
                second_nums.push_back(second);
                first_index++;
                second_index--;
            }
            else if(first + second < sum){
                first_index++;
            }
            else{ // first + second > sum
                second_index--;
            }
        }

        int min_index = 0;
        int min; // 未初始化最小值
        for(int i =0 ;i < first_nums.size(); i++){
            if(first_nums[i]*second_nums[i] < min){
                min_index = i;
                min = first_nums[i]*second_nums[i];
            }
        }

        ret.push_back(first_nums[min_index]);
        ret.push_back(second_nums[min_index]);

        return ret;
    }
};

maybe ok

#include "stdio.h"
#include <vector>

using std::vector;

class Solution {
public:
    vector<int> FindNumbersWithSum(vector<int> array,int sum) {
        vector<int> ret;

        if(array.size()==0){
            return ret;
        };

        vector<int> first_nums;
        vector<int> second_nums;

        int first_index = 0;
        int second_index = array.size() - 1;

        while((first_index < second_index)&&(first_index <array.size()-1)&&(second_index>0)){// 注意索引不要越界。
            int first = array[first_index];
            int second = array[second_index];

            if( first + second == sum) {
                first_nums.push_back(first);
                second_nums.push_back(second);
                first_index++;
                second_index--;
            }
            else if(first + second < sum){
                first_index++;
            }
            else{ // first + second > sum
                second_index--;
            }
        }

        if(first_nums.size()>0){
            int min_index = 0;
            int min = first_nums[0]*second_nums[0];// 1.若是vector爲空可能形成越界，需先判斷vector是否爲空。2. 用於比較的最小值要先賦初始值。

            for(int i =0 ;i < first_nums.size(); i++){
                if(first_nums[i]*second_nums[i] < min){
                    min_index = i;
                    min = first_nums[i]*second_nums[i];
                }
            }

            ret.push_back(first_nums[min_index]);
            ret.push_back(second_nums[min_index]);
        }

        return ret;
    }
};

int main(int argc, char const *argv[])
{
    /* code */
    int nums[] = {1,1,1,3,1};
    std::vector<int> vnums(nums,nums+5);
    Solution sol;
    
    std::vector<int> vret;
    vret = sol.FindNumbersWithSum(vnums,5);

    for(std::vector<int>::iterator it= vret.begin(); it != vret.end();it++){
        printf("%d\n", *it);
    }    
    return 0;

}