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
思路：二分查找
注意：k用来记录中间数的下标。当target!=A[k]后，如何分很关键。接下来的查找将不再考虑元素A[k]，而是在[0,k-1]或[k+1,n-1]中查找。
why？？？
1 class Solution { 2 public: 3 int searchInsert(int A[], int n, int target) { 4 return binarySearch(A,0,n-1,target); 5 } 6 7 int binarySearch(int A[],int a, int b, int target){ 8 if(a>b) return a; 9 int k = (a+b)/2; 10 if(target == A[k]) return k; 11 else if(target<A[k]) return binarySearch(A,0,k-1,target); 12 else return binarySearch(A,k+1,b,target); 13 } 14 };
原文：http://www.cnblogs.com/renrenbinbin/p/4330798.html