版权声明：文章都是原创，转载请注明~~~~ https://blog.csdn.net/SourDumplings/article/details/86529202
Easy
1068154FavoriteShare
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:
Input: [1,3,5,6], 5
Output: 2
Example 2:
Input: [1,3,5,6], 2
Output: 1
Example 3:
Input: [1,3,5,6], 7
Output: 4
Example 4:
Input: [1,3,5,6], 0
Output: 0
C++：
/*
@Date : 2019-01-15 20:52:16
@Author : 酸饺子 ([email protected])
@Link : https://github.com/SourDumplings
@Version : $Id$
*/
/*
https://leetcode.com/problems/search-insert-position/
*/
class Solution
{
public:
int searchInsert(vector<int>& nums, int target)
{
int b = 0, e = nums.size();
int mi;
while (b < e)
{
mi = (b + e) >> 1;
target < nums[mi] ? e = mi : b = mi + 1;
}
if (b == 0 || nums[b - 1] < target)
{
return b;
}
else
return b - 1;
}
};
Java：
/**
* @Date : 2019-01-15 21:24:53
* @Author : 酸饺子 ([email protected])
* @Link : https://github.com/SourDumplings
* @Version : $Id$
*
* https://leetcode.com/problems/search-insert-position/
*/
class Solution
{
public int searchInsert(int[] nums, int target)
{
int b = 0, e = nums.length;
while (b < e)
{
int mi = (b + e) >> 1;
if (target < nums[mi])
e = mi;
else
b = mi + 1;
}
if (b == 0 || nums[b - 1] < target)
{
return b;
}
else
return b - 1;
}
}