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

基本都是用二分法吧，只是在while循环前先将边界情况考虑，while循环的结束条件是left<right-1，这样当最后一轮循环后left=mid,right=left+1;

若mid==target,返回mid;否则返回right。(若right==target正好，若right<target，则right即为target将要插入的位置)

代码：

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