[Leetcode] 683. K Empty Slots 解题报告

题目：

There is a garden with 

N

 slots. In each slot, there is a flower. The 

N

 flowers
will bloom one by one in 

N

 days. In each day, there will be 

exactly

 one
flower blooming and it will be in the status of blooming since then.

Given an array 

flowers

 consists of number from 

1

 to 

N

.
Each number in the array represents the place where the flower will open in that day.

For example, 

flowers[i] = x

 means that the unique flower that blooms at day 

i

 will
be at position 

x

, where 

i

 and 

x

 will
be in the range from 

1

 to 

N

.

Also given an integer 

k

, you need to output in which day there exists two flowers in the
status of blooming, and also the number of flowers between them is 

k

 and these flowers
are not blooming.

If there isn't such day, output -1.

Example 1:

Input:
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.

Example 2:

Input:
flowers: [1,2,3]
k: 1
Output: -1

Note:

The given array will be in the range [1, 20000].
思路：

本题目描述中的数组索引是从1开始的，这和程序猿们通常让索引从0开始的习惯有所出入。为了统一，我们在描述中对于索引还是从1开始，只是在实现中稍加改变。

思路是：我们额外定义一个数组days，用来表示每朵花的开花时间，例如days[i] = y表示在位置i上的花的开花时间是第y天。那么问题就转化为：给定days数组之后，找出一个子区间[left, left + 1, ...left + k - 1, right)，满足对于任意的i = left + 1,...left + k - 1，满足days[left] < days[i] && days[right] < days[i]。如果找到了，就返回true，否则就返回false。

代码：

class Solution {
public:
int kEmptySlots(vector<int>& flowers, int k) {
vector<int> days(flowers.size());
for(int i = 0; i < flowers.size(); ++i) {
days[flowers[i] - 1] = i + 1;
}
int left = 0, right = k + 1, res = INT_MAX;
for(int i = 0; right < days.size(); ++i) {
if(days[i] < days[left] || days[i] <= days[right]) {
if(i == right) {
res = min(res, max(days[left], days[right])); // we get a valid subarray
}
left = i, right = k + 1 + i;
}
}
retur
a920
n (res == INT_MAX)? -1 : res;
}
};