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feat: update lc problems #3892

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Dec 27, 2024
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feat: update lc problems #3892

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124 changes: 62 additions & 62 deletions solution/0000-0099/0011.Container With Most Water/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -62,13 +62,15 @@ tags:

### 方法一：双指针

一开始，我们考虑相距最远的两个柱子所能容纳水的容量。水的宽度是两根柱子之间的距离，而水的高度取决于两根柱子之间较短的那个。
我们使用两个指针 $l$ 和 $r$ 分别指向数组的左右两端，即 $l = 0$，而 $r = n - 1$，其中 $n$ 是数组的长度。

当前柱子是最两侧的柱子，水的宽度最大，其他的组合，水的宽度都比这个小。不妨假设左侧柱子的高度小于等于右侧柱子的高度，那么水的高度就是左侧柱子的高度。如果我们移动右侧柱子，那么水的宽度就减小了，而水的高度却不会增加，因此水的容量一定减少。所以我们移动左侧柱子，更新最大容量。
接下来，我们使用变量 $\textit{ans}$ 记录容器的最大容量，初始化为 $0$。

循环此过程，直到两个柱子相遇。
然后，我们开始进行循环，每次循环中，我们计算当前容器的容量，即 $\textit{min}(height[l], height[r]) \times (r - l)$，并将其与 $\textit{ans}$ 进行比较，将较大值赋给 $\textit{ans}$。然后，我们判断 $height[l]$ 和 $height[r]$ 的大小，如果 $\textit{height}[l] < \textit{height}[r]$，移动 $r$ 指针不会使得结果变得更好，因为容器的高度由较短的那根垂直线决定，所以我们移动 $l$ 指针。反之，我们移动 $r$ 指针。

时间复杂度 $O(n)$，其中 $n$ 是数组 `height` 的长度。空间复杂度 $O(1)$。
遍历结束后，返回 $\textit{ans}$ 即可。

时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{height}$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -77,15 +79,15 @@ tags:
```python
class Solution:
def maxArea(self, height: List[int]) -> int:
i, j = 0, len(height) - 1
l, r = 0, len(height) - 1
ans = 0
while i < j:
t = (j - i) * min(height[i], height[j])
while l < r:
t = min(height[l], height[r]) * (r - l)
ans = max(ans, t)
if height[i] < height[j]:
i += 1
if height[l] < height[r]:
l += 1
else:
j -= 1
r -= 1
return ans
```

Expand All @@ -94,15 +96,15 @@ class Solution:
```java
class Solution {
public int maxArea(int[] height) {
int i = 0, j = height.length - 1;
int l = 0, r = height.length - 1;
int ans = 0;
while (i < j) {
int t = Math.min(height[i], height[j]) * (j - i);
while (l < r) {
int t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[i] < height[j]) {
++i;
if (height[l] < height[r]) {
++l;
} else {
--j;
--r;
}
}
return ans;
Expand All @@ -116,15 +118,15 @@ class Solution {
class Solution {
public:
int maxArea(vector<int>& height) {
int i = 0, j = height.size() - 1;
int l = 0, r = height.size() - 1;
int ans = 0;
while (i < j) {
int t = min(height[i], height[j]) * (j - i);
while (l < r) {
int t = min(height[l], height[r]) * (r - l);
ans = max(ans, t);
if (height[i] < height[j]) {
++i;
if (height[l] < height[r]) {
++l;
} else {
--j;
--r;
}
}
return ans;
Expand All @@ -136,14 +138,14 @@ public:

```go
func maxArea(height []int) (ans int) {
i, j := 0, len(height)-1
for i < j {
t := min(height[i], height[j]) * (j - i)
l, r := 0, len(height)-1
for l < r {
t := min(height[l], height[r]) * (r - l)
ans = max(ans, t)
if height[i] < height[j] {
i++
if height[l] < height[r] {
l++
} else {
j--
r--
}
}
return
Expand All @@ -154,16 +156,15 @@ func maxArea(height []int) (ans int) {

```ts
function maxArea(height: number[]): number {
let i = 0;
let j = height.length - 1;
let [l, r] = [0, height.length - 1];
let ans = 0;
while (i < j) {
const t = Math.min(height[i], height[j]) * (j - i);
while (l < r) {
const t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[i] < height[j]) {
++i;
if (height[l] < height[r]) {
++l;
} else {
--j;
--r;
}
}
return ans;
Expand All @@ -175,15 +176,15 @@ function maxArea(height: number[]): number {
```rust
impl Solution {
pub fn max_area(height: Vec<i32>) -> i32 {
let mut i = 0;
let mut j = height.len() - 1;
let mut l = 0;
let mut r = height.len() - 1;
let mut ans = 0;
while i < j {
ans = ans.max(height[i].min(height[j]) * ((j - i) as i32));
if height[i] <= height[j] {
i += 1;
while l < r {
ans = ans.max(height[l].min(height[r]) * ((r - l) as i32));
if height[l] < height[r] {
l += 1;
} else {
j -= 1;
r -= 1;
}
}
ans
Expand All @@ -199,16 +200,15 @@ impl Solution {
* @return {number}
*/
var maxArea = function (height) {
let i = 0;
let j = height.length - 1;
let [l, r] = [0, height.length - 1];
let ans = 0;
while (i < j) {
const t = Math.min(height[i], height[j]) * (j - i);
while (l < r) {
const t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[i] < height[j]) {
++i;
if (height[l] < height[r]) {
++l;
} else {
--j;
--r;
}
}
return ans;
Expand All @@ -220,15 +220,15 @@ var maxArea = function (height) {
```cs
public class Solution {
public int MaxArea(int[] height) {
int i = 0, j = height.Length - 1;
int l = 0, r = height.Length - 1;
int ans = 0;
while (i < j) {
int t = Math.Min(height[i], height[j]) * (j - i);
while (l < r) {
int t = Math.Min(height[l], height[r]) * (r - l);
ans = Math.Max(ans, t);
if (height[i] < height[j]) {
++i;
if (height[l] < height[r]) {
++l;
} else {
--j;
--r;
}
}
return ans;
Expand All @@ -245,16 +245,16 @@ class Solution {
* @return Integer
*/
function maxArea($height) {
$i = 0;
$j = count($height) - 1;
$l = 0;
$r = count($height) - 1;
$ans = 0;
while ($i < $j) {
$t = min($height[$i], $height[$j]) * ($j - $i);
while ($l < $r) {
$t = min($height[$l], $height[$r]) * ($r - $l);
$ans = max($ans, $t);
if ($height[$i] < $height[$j]) {
++$i;
if ($height[$l] < $height[$r]) {
++$l;
} else {
--$j;
--$r;
}
}
return $ans;
Expand Down
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