2517.礼盒的最大甜蜜度
链接:2517.礼盒的最大甜蜜度
难度:Medium
标签:贪心、数组、二分查找、排序
简介:返回礼盒的 最大 甜蜜度。
题解 1 - cpp
- 编辑时间:2022-12-25
- 执行用时:172ms
- 内存消耗:49.4MB
- 编程语言:cpp
- 解法介绍:二分答案。
class Solution {
public:
int n, k;
vector<int> price;
int maximumTastiness(vector<int>& price, int k) {
sort(price.begin(), price.end());
n = price.size();
this->k = k;
this->price = price;
int l = 0, r = price[n - 1] - price[0], m;
while (l < r) {
m = (l + r + 1) / 2;
if (check(m)) l = m;
else r = m - 1;
}
return l;
}
bool check(int num) {
int cnt = 1, cur = price[0];
for (int i = 1; i < n; i++) {
if (cur + num <= price[i]) {
cnt++;
cur = price[i];
}
}
return cnt >= k;
}
};
题解 2 - cpp
- 编辑时间:2023-06-01
- 执行用时:212ms
- 内存消耗:47.3MB
- 编程语言:cpp
- 解法介绍:二分答案,尽可能找差超过target的数量。
class Solution {
public:
int maximumTastiness(vector<int>& price, int k) {
sort(price.begin(), price.end());
int n = price.size(), l = 0, r = price[n - 1] - price[0];
while (l < r) {
int m = (l + r + 1) / 2, cnt = 1, prev = price[0];
for (int i = 1; i < n; i++) {
if (price[i] - prev >= m) cnt++, prev = price[i];
}
if (cnt < k) r = m - 1;
else l = m;
}
return l;
}
};
题解 3 - python
- 编辑时间:2023-06-01
- 执行用时:996ms
- 内存消耗:27.5MB
- 编程语言:python
- 解法介绍:同上。
class Solution:
def maximumTastiness(self, price: List[int], k: int) -> int:
price.sort()
n = len(price)
l = 0
r = price[n-1]-price[0]
while l < r:
m = (l+r+1)//2
cnt = 1
prev = price[0]
for i in range(1, n):
if price[i] - prev >= m:
cnt += 1
prev = price[i]
if cnt < k:
r = m-1
else:
l = m
return l
题解 4 - rust
- 编辑时间:2023-06-01
- 执行用时:44ms
- 内存消耗:3.9MB
- 编程语言:rust
- 解法介绍:同上。
impl Solution {
pub fn maximum_tastiness(mut price: Vec<i32>, k: i32) -> i32 {
price.sort();
let n = price.len();
let mut l = 0;
let mut r = price[n - 1] - price[0];
while l < r {
let m = (l + r + 1) / 2;
let mut cnt = 1;
let mut prev = price[0];
for i in 1..n {
if price[i] - prev >= m {
cnt += 1;
prev = price[i];
}
}
if cnt < k {
r = m - 1;
} else {
l = m
}
}
l
}
}