leetcode之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

拿到这道题之后，首先第一反应是用二分法，因为是在有序数组中查找某个元素的位置。

根据题目给出的测试用例，非常全的考虑了边界的情况等。这道题一遍编译通过，我也很奇怪。只是有一点不是很明白的地方在于，我的每一个判断语句中有return了，为什么整个函数最后还要加一个return，最后的这个return是返回的什么？不是很清楚。 

下面附上我的代码，运行时间有点长

public class Solution {
    public int searchInsert(int[] A, int target) {
        int result = 0;
        int length = A.length;
        int low = 0;
        int high = length-1;
        while(low<high){
        	int mid = (high+low)/2;
        	if(A[mid]<target){
        		low = mid+1;
        	}
        	else{
        		if(A[mid]>target){
            		high = mid-1;
            	}
        		else{
        			result = mid;
        			return result;
        		}
        	} 
        }
        if(A[low]<target){
        	result = high + 1;
        	return result;
        }
        if(A[low]>target){
        	result = low;
        	return result;
        }
		return low;
       
    }
}