15算法课程 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

solution:

二分查找，当找不到时l=r+1,所以根据最后一次l和r的变动来判定应该插入的位置，如果最后一次是l=mid+1，说明应该插入到mid+1的位置，如果最后一次是r=mid-1，则说明应该插入到mid的位置。

code:

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