#60 Search Insert Position

题目描述：

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.

Have you met this question in a real interview? 

Yes

Example

[1,3,5,6]

, 5 → 2

[1,3,5,6]

, 2 → 1

[1,3,5,6]

, 7 → 4

[1,3,5,6]

, 0 → 0

Challenge 

O(log(n)) time

题目思路：

normal binary search过后，如果没有找到target，l和r可能停在> target或者< target的位置上：如果这个位置<target，那么需要return的是位置+1（因为要insert在后面）；反之，则return那个位置（insert在前面的意思就是代替当前位置）。

Mycode（AC = 35ms）：

class Solution {
/**
* param A : an integer sorted array
* param target : an integer to be inserted
* return : an integer
*/
public:
int searchInsert(vector<int> &A, int target) {
// write your code here
if (A.size() == 0) return 0;

// try to find the target
int l = 0, r = A.size() - 1;
while (l + 1 < r) {
int mid = (l + r) / 2;

if (A[mid] == target) {
return mid;
}
else if (A[mid] < target) {
l = mid;
}
else {
r = mid;
}
}

// there are two possible positions to insert: r & l
if (A[r] < target) {
return r + 1;
}
else if (A[l] < target) {
return l + 1;
}
else {
return l;
}
}
};