1124.表现良好的最长时间段
链接:1124.表现良好的最长时间段
难度:Medium
标签:栈、数组、哈希表、前缀和、单调栈
简介:给你一份工作时间表 hours,上面记录着某一位员工每天的工作小时数。
题解 1 - typescript
- 编辑时间:2021-03-20
- 执行用时:1044ms
- 内存消耗:42.8MB
- 编程语言:typescript
- 解法介绍:前缀和。
function longestWPI(hours: number[]): number {
hours = hours.map(v => (v > 8 ? 1 : -1));
const len = hours.length;
const sums = [0];
let sum = 0;
hours.forEach(hour => sums.push((sum += hour)));
let max = 0;
for (let i = 1; i < len + 1; i++) {
for (let j = 0; j < i; j++) {
if (sums[i] - sums[j] > 0) max = Math.max(max, i - j);
}
}
return max;
}
题解 2 - cpp
- 编辑时间:2023-02-14
- 执行用时:32ms
- 内存消耗:23.3MB
- 编程语言:cpp
- 解法介绍:单调栈,找出最远最小的值。
class Solution {
public:
int longestWPI(vector<int>& hours) {
int n = hours.size(), ans = 0;
for (auto &h : hours) h = h > 8 ? 1 : -1;
vector<int> sums(1, 0);
for (auto &h : hours) sums.push_back(sums.back() + h);
stack<int> s; s.push(0);
for (int i = 1; i <= n; i++) {
if (sums[s.top()] > sums[i]) s.push(i);
}
for (int i = n; i >= 1; i--) {
while (s.size() && sums[s.top()] < sums[i]) {
ans = max(ans, i - s.top());
s.pop();
}
}
return ans;
}
};
题解 3 - python
- 编辑时间:2023-02-14
- 执行用时:72ms
- 内存消耗:15.7MB
- 编程语言:python
- 解法介绍:同上。
class Solution:
def longestWPI(self, hours: List[int]) -> int:
n = len(hours)
ans = 0
sums = [0]
for h in hours:
v = -1
if (h > 8):
v = 1
sums.append(sums[-1] + v)
s = [0]
for i in range(1, n+1):
if sums[s[-1]] > sums[i]:
s.append(i)
for i in range(n, 0, -1):
while len(s) and sums[s[-1]] < sums[i]:
ans = max(ans, i - s.pop())
return ans
题解 4 - rust
- 编辑时间:2023-02-14
- 执行用时:12ms
- 内存消耗:2.4MB
- 编程语言:rust
- 解法介绍:同上。
impl Solution {
pub fn longest_wpi(hours: Vec<i32>) -> i32 {
use std::collections::VecDeque;
let n = hours.len();
let mut ans = 0;
let mut sums = vec![0; 1];
for h in hours {
let h: i32 = if h > 8 { 1 } else { -1 };
sums.push(sums.last().unwrap() + h);
}
let mut s = VecDeque::<usize>::new();
s.push_back(0);
for i in 1..=n {
if sums[*s.back().unwrap()] > sums[i] {
s.push_back(i);
}
}
let mut i = n;
while i >= 1 {
while !s.is_empty() && sums[*s.back().unwrap()] < sums[i] {
ans = ans.max(i as i32 - s.pop_back().unwrap() as i32);
}
i -= 1;
}
ans
}
}