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
本题是基本考察二分查找的题目，与基本二分查找方法不同的地方是，二分查找法当查找的target不在list中存在时返回-1，而本题则需要返回该target应在此list中插入的位置。
当循环结束时，如果没有找到target，那么low一定停target应该插入的位置上，high一定停在恰好比target小的index上。
[1,3,5,6], 7
low = 0, high = 3
step1: mid = 1
A[mid] = 3, 3<7
low = mid + 1 = 2
low = 2, high = 3
step2: mid = 2
A[mid] = 5, 5<7
low = mid + 1 = 3
low = 3, high = 3
step3: mid = 3
A[mid] = 6, 6<7
low = mid + 1 = 4
low = 4, high = 3
return low = 4;
[1,3,5,6], 0
low = 0, high = 3
step1: mid = 1
A[mid] = 3, 3>0
high = mid - 1 = 0
low = 0, high = 0
step2: mid = 0
A[mid] = 1, 1>0
high = mid - 1 = -1
low = 0, high = -1
return 0
2 if(A == null||A.length == 0)
3 return 0;
4 int low = 0, high = A.length-1;
5
6 while(low <= high){
7 int mid = (low + high) / 2;
8
9 if(A[mid] > target)
10 high = mid - 1;
11 else if(A[mid] < target)
12 low = mid + 1;
13 else
14 return mid;
15 }
16
17 return low;
18 }
自己在做第二遍时候犯二，mid = (low+high)/2的括号忘记加了
Reference: http://blog.csdn.net/linhuanmars/article/details/31354941