Question

剑指 Offer II 039. 直方图最大矩形面积

题目描述

给定非负整数数组 heights ，数组中的数字用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。

求在该柱状图中，能够勾勒出来的矩形的最大面积。

 

示例 1:

输入：heights = [2,1,5,6,2,3]
输出：10
解释：最大的矩形为图中红色区域，面积为 10

示例 2：

输入： heights = [2,4]
输出： 4

 

提示：

• 1 <= heights.length <=105
• 0 <= heights[i] <= 104

 

注意：本题与主站 84 题相同： https://leetcode.cn/problems/largest-rectangle-in-histogram/

解法

方法一：单调栈

单调栈常见模型：找出每个数左/右边离它最近的且比它大/小的数。模板：

stk = []
for i in range(n):
  while stk and check(stk[-1], i):
    stk.pop()
  stk.append(i)

枚举每根柱子的高度 $h$ 作为矩形的高度，向左右两边找第一个高度小于 $h$ 的下标 $left_i$, $right_i$。那么此时矩形面积为 $h \times (right_i-left_i-1)$，求最大值即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为柱子个数。

class Solution:
  def largestRectangleArea(self, heights: List[int]) -> int:
    n = len(heights)
    left = [-1] * n
    right = [n] * n
    stk = []
    for i, x in enumerate(heights):
      while stk and heights[stk[-1]] >= x:
        stk.pop()
      if stk:
        left[i] = stk[-1]
      stk.append(i)
    stk = []
    for i in range(n - 1, -1, -1):
      while stk and heights[stk[-1]] >= heights[i]:
        stk.pop()
      if stk:
        right[i] = stk[-1]
      stk.append(i)
    return max(x * (r - l - 1) for x, l, r in zip(heights, left, right))

class Solution {
  public int largestRectangleArea(int[] heights) {
    int n = heights.length;
    int[] left = new int[n];
    int[] right = new int[n];
    for (int i = 0; i < n; ++i) {
      left[i] = -1;
      right[i] = n;
    }
    Deque<Integer> stk = new ArrayDeque<>();
    for (int i = 0; i < n; ++i) {
      while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
        stk.pop();
      }
      if (!stk.isEmpty()) {
        left[i] = stk.peek();
      }
      stk.push(i);
    }
    stk.clear();
    for (int i = n - 1; i >= 0; --i) {
      while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
        stk.pop();
      }
      if (!stk.isEmpty()) {
        right[i] = stk.peek();
      }
      stk.push(i);
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      ans = Math.max(ans, (right[i] - left[i] - 1) * heights[i]);
    }
    return ans;
  }
}

class Solution {
public:
  int largestRectangleArea(vector<int>& heights) {
    int n = heights.size();
    vector<int> left(n, -1), right(n, n);
    stack<int> stk;
    for (int i = 0; i < n; ++i) {
      while (!stk.empty() && heights[stk.top()] >= heights[i]) {
        stk.pop();
      }
      if (!stk.empty()) {
        left[i] = stk.top();
      }
      stk.push(i);
    }
    stk = stack<int>();
    for (int i = n - 1; ~i; --i) {
      while (!stk.empty() && heights[stk.top()] >= heights[i]) {
        stk.pop();
      }
      if (!stk.empty()) {
        right[i] = stk.top();
      }
      stk.push(i);
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      ans = max(ans, (right[i] - left[i] - 1) * heights[i]);
    }
    return ans;
  }
};

func largestRectangleArea(heights []int) (ans int) {
  n := len(heights)
  left := make([]int, n)
  right := make([]int, n)
  for i := range left {
    left[i] = -1
    right[i] = n
  }
  stk := []int{}
  for i, x := range heights {
    for len(stk) > 0 && heights[stk[len(stk)-1]] >= x {
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      left[i] = stk[len(stk)-1]
    }
    stk = append(stk, i)
  }
  stk = []int{}
  for i := n - 1; i >= 0; i-- {
    for len(stk) > 0 && heights[stk[len(stk)-1]] >= heights[i] {
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      right[i] = stk[len(stk)-1]
    }
    stk = append(stk, i)
  }
  for i, x := range heights {
    ans = max(ans, (right[i]-left[i]-1)*x)
  }
  return
}

function largestRectangleArea(heights: number[]): number {
  const n = heights.length;
  const left: number[] = new Array(n).fill(-1);
  const right: number[] = new Array(n).fill(n);
  const stk: number[] = [];
  for (let i = 0; i < n; ++i) {
    while (stk.length && heights[stk[stk.length - 1]] >= heights[i]) {
      stk.pop();
    }
    if (stk.length) {
      left[i] = stk[stk.length - 1];
    }
    stk.push(i);
  }
  stk.length = 0;
  for (let i = n - 1; i >= 0; --i) {
    while (stk.length && heights[stk[stk.length - 1]] >= heights[i]) {
      stk.pop();
    }
    if (stk.length) {
      right[i] = stk[stk.length - 1];
    }
    stk.push(i);
  }
  let ans = 0;
  for (let i = 0; i < n; ++i) {
    ans = Math.max(ans, (right[i] - left[i] - 1) * heights[i]);
  }
  return ans;
}