LeetCode @ Search Insert Position 二分查找

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

<span style="font-size:12px;">//来源于Code_Ganker:http://blog.csdn.net/linhuanmars/article/details/20278967 //"简单的二分：思路就是每次取中间，如果等于目标即返回；否则根据大小关系切去一半。因此算法复杂度是O(logn)，空间复杂度O(1)。public class Solution {    public int searchInsert(int[] A, int target) {        int left=0;        int right=A.length-1;        while(left<=right){//若找不到目标，最后一次操作是left和right指向同一index，执行一次循环体；结束时left>right，左右指针位置颠倒。            int mid=(left+right)/2;            if(target==A[mid])                return mid;            if (target<A[mid])//left half                right=mid-1;            else  //right half                left=mid+1;        }        return left;//"注意以上实现有一个好处:循环结束时，如果没找到目标，那么left一定停在恰好比目标大的index上，right一定停在恰好比目标小的index上，比较推荐这种实现。"    }}</span>

注：1.《编程之美》上关于二分法mid变量的处理

2.会写递归二分法