【LeetCode】Search Insert Position

Search Insert Position
Total Accepted: 10088 Total Submissions: 29566 My Submissions
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
这个应该是二分查找吧。
注意两点：1、如果数组为空，直接返回0。2、如果target比A[0]小或者比A[len-1]大，返回0或者Len即可。
然后查找，如果low <= high的话，无论找不找得到，都直接范围index。

Java AC 1 

public class Solution {
    public int searchInsert(int[] A, int target) {
        if(A == null || A.length == 0){
            return 0;
        }
        int len = A.length;
        if (target < A[0]) {
			return 0;
		}
        if (target > A[len-1]) {
			return len;
		}
        int low = 0;
        int high = len - 1;
        int mid = 0;
        while(low <= high){
            mid = (low + high) >> 1;
            if(A[mid] > target){
                high = mid - 1;
            }else if(A[mid] < target){
                low = mid + 1;
            }else{
                return mid;
            }
        }
		return low;
    }
}

Java AC 2 其实搜索过程应该是1要快些，但是LeetCode好像这个快了将近20ms，不应该呀。

public class Solution {
    public int searchInsert(int[] A, int target) {
        if(A == null || A.length == 0){
            return 0;
        }
        int len = A.length;
        if (target < A[0]) {
			return 0;
		}
        if (target > A[len-1]) {
			return len;
		}
        int low = 0;
        int high = len - 1;
        int mid = 0;
        while(low <= high){
            mid = (low + high) >> 1;
            if(A[mid] > target){
                high--;
            }else if(A[mid] < target){
                low++;
            }else{
                break;
            }
        }
		return mid;
    }
}