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:

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

C语言源代码（7ms）：

int searchInsert(int* nums, int numsSize, int target) {
    int l=0,r=numsSize-1,pos=0,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(nums[mid]==target)return mid;
        else if(target>nums[mid]){
            l=mid+1;pos=mid+1;
        }else{
            r=mid-1;
            pos=mid;
        } 
    }
    return pos;
}

C++源代码（8ms）：

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;
    }
};

Python源代码（52ms）：

class Solution:
    # @param {integer[]} nums
    # @param {integer} target
    # @return {integer}
    def searchInsert(self, nums, target):
        l=0;r=len(nums)-1;pos=0
        while l<=r:
            mid=(l+r)>>1
            if target==nums[mid]:return mid
            elif target>nums[mid]:l=mid+1;pos=mid+1
            else:r=mid-1;pos=mid
        return pos