Search Insert Position

问题描述：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], 7→ 4
[1,3,5,6], 2 → 1
[1,3,5,6], 0 → 0

思路：首先想到的就是用遍历一遍解决这个问题，因为题目没有时间的限制，当然更好的办法是利用二分查找的方法。

代码：

class Solution {public:    int searchInsert(vector<int>& nums, int target) {        if(nums.size() < 1) return 0;        if(target <= nums[0]) return 0;        for(int i = 0; i < nums.size(); ++i){            if(nums[i] == target) return i;            if(nums[i] < target && target < nums[i+1] && i < nums.size()-1)                return i+1;        }        if(nums[nums.size()-1] < target) return nums.size();    }};

运行时间：O(n)