[pull] master from youngyangyang04:master

problems/0042.接雨水.md

@@ -926,6 +926,56 @@ int trap(int* height, int heightSize) {
 * 空间复杂度 O(1)
 
 
+Rust
+
+双指针
+
+```rust
+impl Solution {
+    pub fn trap(height: Vec<i32>) -> i32 {
+        let n = height.len();
+        let mut max_left = vec![0; height.len()];
+        let mut max_right = vec![0; height.len()];
+        max_left.iter_mut().zip(max_right.iter_mut().rev()).enumerate().fold((0, 0), |(lm, rm), (idx, (x, y))| {
+            let lmax = lm.max(height[idx]);
+            let rmax = rm.max(height[n - 1 - idx]);
+            *x = lmax; *y = rmax;
+            (lmax, rmax)
+        });
+        height.iter().enumerate().fold(0, |acc, (idx, x)| {
+            let h = max_left[idx].min(max_right[idx]);
+            if h > 0 { h - x + acc } else { acc }
+        })
+    }
+}
+```
+
+单调栈
+
+```rust
+impl Solution {
+    pub fn trap(height: Vec<i32>) -> i32 {
+        let mut stack = vec![];
+        let mut ans = 0;
+        for (right_pos, &right_h) in height.iter().enumerate() {
+            while !stack.is_empty() && height[*stack.last().unwrap()] <= right_h {
+                let mid_pos = stack.pop().unwrap();
+                if !stack.is_empty() {
+                    let left_pos = *stack.last().unwrap();
+                    let left_h = height[left_pos];
+                    let top = std::cmp::min(left_h, right_h);
+                    if top > height[mid_pos] {
+                        ans += (top - height[mid_pos]) * (right_pos - left_pos - 1) as i32;
+                    }
+                }
+            }
+            stack.push(right_pos);
+        }
+        ans
+    }
+}
+```
+
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problems/0084.柱状图中最大的矩形.md

@@ -670,6 +670,61 @@ function largestRectangleArea(heights: number[]): number {
 ```
 
 
+Rust
+
+双指针预处理
+```rust
+
+impl Solution {
+    pub fn largest_rectangle_area(v: Vec<i32>) -> i32 {
+        let n = v.len();
+        let mut left_smaller_idx = vec![-1; n];
+        let mut right_smaller_idx = vec![n as i32; n];
+        for i in 1..n {
+            let mut mid = i as i32 - 1;
+            while mid >= 0 && v[mid as usize] >= v[i] {
+                mid = left_smaller_idx[mid as usize];
+            }
+            left_smaller_idx[i] = mid;
+        }
+        for i in (0..n-1).rev() {
+            let mut mid = i + 1;
+            while mid < n && v[mid] >= v[i] {
+                mid = right_smaller_idx[mid] as usize;
+            }
+            right_smaller_idx[i] = mid as i32;
+        }
+        let mut res = 0;
+        for (idx, &e) in v.iter().enumerate() {
+            res = res.max((right_smaller_idx[idx] - left_smaller_idx[idx] - 1) * e);
+        }
+        dbg!(res)
+    }
+}
+```
+
+单调栈
+```rust
+impl Solution {
+    pub fn largest_rectangle_area1(mut v: Vec<i32>) -> i32 {
+        v.insert(0, 0);  // 便于使第一个元素能够有左侧<=它的值
+        v.push(0);                      // 便于在结束处理最后一个元素后清空残留在栈中的值
+        let mut res = 0;
+        let mut stack = vec![]; // 递增的栈
+        for (idx, &e) in v.iter().enumerate() {
+            while !stack.is_empty() && v[*stack.last().unwrap()] > e {
+                let pos = stack.pop().unwrap();
+                let prev_pos = *stack.last().unwrap();
+                let s = (idx - prev_pos - 1) as i32 * v[pos];
+                res = res.max(s);
+            }
+            stack.push(idx);
+        }
+        res
+    }  
+}
+```
+
 
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problems/0503.下一个更大元素II.md

@@ -266,6 +266,24 @@ function nextGreaterElements(nums: number[]): number[] {
 };
 ```
 
+Rust
+```rust
+impl Solution {
+    pub fn next_greater_elements(nums: Vec<i32>) -> Vec<i32> {
+        let mut ans = vec![-1; nums.len() * 2];
+        let mut stack = vec![];
+        let double = nums.repeat(2);
+        for (idx, &i) in double.iter().enumerate() {
+            while !stack.is_empty() && double[*stack.last().unwrap()] < i {
+                let pos = stack.pop().unwrap();
+                ans[pos] = i;
+            }
+            stack.push(idx);
+        }
+        ans.into_iter().take(nums.len()).collect()
+    }
+}
+```
 
 
 <p align="center">