No.35 Search Insert Position

No.35 Search Insert Position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

---

#include "stdafx.h"
#include <vector>
#include <iostream>
using namespace std;

class Solution
{
public:
    int searchInsert(vector<int> & numbers, int target)
    {//典型的二分查找
     //输入：数组numbers(无重复元素)，target待查找的元素
     //输出：若target在numbers中，则输出其下标(从0开始)；否则，输出若target插入数组，在数组中的下标
        
        //判断是否可查，即数组是否为空
        if(numbers.size() == 0)
            return 0;//此时若插入，下标为0
        int low = 0;
        int height = numbers.size()-1;
        int middle;

        while(low <= height)
        {
            middle = (low+height)/2;
            if(numbers[middle] == target)
                return middle;
            else
                if(numbers[middle] < target)
                    low = middle + 1;
                else
                    height = middle - 1;
        }
        if(low != height)
            return (low > height ? low : height);
        //return low;
        //当循环结束时，如果没有找到目标元素，那么low一定停在恰好比目标大的index上，height一定停在恰好比目标小的index上
    }
};

int main()
{
    Solution sol;
    int data[] = {1,3,5,6};
    vector<int> test(data,data+sizeof(data)/sizeof(int));

    cout <<"5 -> "<< sol.searchInsert(test, 5)<< endl;
    cout <<"2 -> "<< sol.searchInsert(test, 2)<< endl;
    cout <<"7 -> "<< sol.searchInsert(test, 7)<< endl;
    cout <<"0 -> "<< sol.searchInsert(test, 0)<< endl;
}