# LeetCode 1465. 切割后面积最大的蛋糕

horizontalCuts[i] 是从矩形蛋糕顶部到第 i 个水平切口的距离

verticalCuts[j] 是从矩形蛋糕的左侧到第 j 个竖直切口的距离

输入：h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]

输入：h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]

输入：h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]

2 <= h, w <= 109
1 <= horizontalCuts.length <= min(h - 1, 105)
1 <= verticalCuts.length <= min(w - 1, 105)
1 <= horizontalCuts[i] < h
1 <= verticalCuts[i] < w

horizontalCuts\textit{horizontalCuts}horizontalCuts，和竖直切割方案 verticalCuts\textit{verticalCuts}verticalCuts 对蛋糕进行切割，其中 horizontalCuts[i]\textit{horizontalCuts}[i]horizontalCuts[i] 表示从矩形蛋糕顶部水平往下距离 horizontalCuts[i]\textit{horizontalCuts}[i]horizontalCuts[i] 的位置进行切割，verticalCuts[j]\textit{verticalCuts}[j]verticalCuts[j] 表示从矩形蛋糕最左侧往右距离 verticalCuts[j]\textit{verticalCuts}[j]verticalCuts[j] 的位置进行切割。现在我们需要求出切割后的面积最大的蛋糕面积对 109+710 ^ 9 + 710 9 +7 取模后的值。

9+7 的取模即可。

### C++

class Solution {
public:
int maxArea(int h, int w, vector<int>& horizontalCuts, vector<int>& verticalCuts) {
horizontalCuts.push_back(0);
horizontalCuts.push_back(h);
verticalCuts.push_back(0);
verticalCuts.push_back(w);
sort(horizontalCuts.begin(), horizontalCuts.end());
sort(verticalCuts.begin(), verticalCuts.end());
int x = 0, y = 0;
for (int i = 1; i < horizontalCuts.size(); ++i) {
x = max(x, horizontalCuts[i] - horizontalCuts[i - 1]);
}
for (int i = 1; i < verticalCuts.size(); ++i) {
y = max(y, verticalCuts[i] - verticalCuts[i - 1]);
}
const int mod = 1e9 + 7;
return (1ll * x * y) % mod;
}
};

### python

class Solution:
def maxArea(
self, h: int, w: int, horizontalCuts: List[int], verticalCuts: List[int]
) -> int:
horizontalCuts.extend([0, h])
verticalCuts.extend([0, w])
horizontalCuts.sort()
verticalCuts.sort()
x = max(b - a for a, b in pairwise(horizontalCuts))
y = max(b - a for a, b in pairwise(verticalCuts))
return (x * y) % (10**9 + 7)

### java

class Solution {
public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
final int mod = (int) 1e9 + 7;
Arrays.sort(horizontalCuts);
Arrays.sort(verticalCuts);
int m = horizontalCuts.length;
int n = verticalCuts.length;
long x = Math.max(horizontalCuts[0], h - horizontalCuts[m - 1]);
long y = Math.max(verticalCuts[0], w - verticalCuts[n - 1]);
for (int i = 1; i < m; ++i) {
x = Math.max(x, horizontalCuts[i] - horizontalCuts[i - 1]);
}
for (int i = 1; i < n; ++i) {
y = Math.max(y, verticalCuts[i] - verticalCuts[i - 1]);
}
return (int) ((x * y) % mod);
}
}

### GO

func maxArea(h int, w int, horizontalCuts []int, verticalCuts []int) int {
horizontalCuts = append(horizontalCuts, []int{0, h}...)
verticalCuts = append(verticalCuts, []int{0, w}...)
sort.Ints(horizontalCuts)
sort.Ints(verticalCuts)
x, y := 0, 0
const mod int = 1e9 + 7
for i := 1; i < len(horizontalCuts); i++ {
x = max(x, horizontalCuts[i]-horizontalCuts[i-1])
}
for i := 1; i < len(verticalCuts); i++ {
y = max(y, verticalCuts[i]-verticalCuts[i-1])
}
return (x * y) % mod
}
func max(a, b int) int {
if a > b {
return a
}
return b
}

### RUST

impl Solution {
pub fn max_area(h: i32, w: i32, mut horizontal_cuts: Vec<i32>, mut vertical_cuts: Vec<i32>) -> i32 {
const MOD: i64 = 1_000_000_007;
horizontal_cuts.sort();
vertical_cuts.sort();
let m = horizontal_cuts.len();
let n = vertical_cuts.len();
let mut x = i64::max(horizontal_cuts[0] as i64, h as i64 - horizontal_cuts[m - 1] as i64);
let mut y = i64::max(vertical_cuts[0] as i64, w as i64 - vertical_cuts[n - 1] as i64);
for i in 1..m {
x = i64::max(x, horizontal_cuts[i] as i64 - horizontal_cuts[i - 1] as i64);
}
for i in 1..n {
y = i64::max(y, vertical_cuts[i] as i64 - vertical_cuts[i - 1] as i64);
}
((x * y) % MOD) as i32
}
}

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