【简单】Lintcode 60: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.

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

解题思路1：

1、很容易看出，遍历一次A数组，应该插入target <= A[i] 第一次出现的位置，当全部遍历还未有结果则返回A.size()表示应插入末尾。代码实现很容易，但时间复杂度为O(n).

2、若想有更优复杂度，则需要用二分查找法，见解题思路2。

class Solution {
public:
/*
* @param A: an integer sorted array
* @param target: an integer to be inserted
* @return: An integer
*/
int searchInsert(vector<int> &A, int target)
{
// write your code here

for(int i=0;i<A.size();i++)
{
if(target <= A[i])
return i;
}

return A.size();
}
};解题思路2：
1、在使用二分查找之前，要先对特殊情况进行处理，当target<A[0]时返回0，因为找不到target时，我们采用返回值是left，它是一直向右移动，不能返回0，所以提前考虑这种情况。

class Solution {
public:
/*
* @param A: an integer sorted array
* @param target: an integer to be inserted
* @return: An integer
*/
int searchInsert(vector<int> &A, int target)
{
int left = 0;
int right = A.size() - 1;
int mid = (right + left) / 2;

if (A.size()
4000
!= 0 && *A.begin() > target)
return 0;

while (left <= right)
{
if (A[mid] == target)
return mid;
else if (A[mid] > target)
right = mid - 1;
else
left = mid + 1;

mid = (right + left) / 2;
}

return left;
}
};