Given n non-negative integers, they are used to represent the height of each column in the histogram. Each pillar is adjacent to each other and has a width of 1.
Find the maximum area of the rectangle that can be outlined in the histogram.
The above is an example of a histogram, where the width of each column is 1, and the given height is [2,1,5,6,2,3].
The shaded part in the figure is the largest rectangular area that can be outlined, and its area is 10 units.
Example:
Input: [2,1,5,6,2,3]
Output: 10
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/largest-rectangle-in-histogram The
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Idea: I only thought of the violent solution O(n^2) from left to right for this question. I posted it directly after reading the official O(n). The official solution also has two violent solutions, fixed high and fixed Left node.
ps: It doesn’t seem to be sent, send it again
class Solution {
public int largestRectangleArea(int[] heights) {
int n = heights.length;
int[] left = new int[n];
int[] right = new int[n];
Stack<Integer> mono_stack = new Stack<Integer>();
for (int i = 0; i < n; ++i) {
while (!mono_stack.isEmpty() && heights[mono_stack.peek()] >= heights[i]) {
mono_stack.pop();
}
//最近的小于其的左边的位置
left[i] = (mono_stack.isEmpty() ? -1 : mono_stack.peek());
mono_stack.push(i);
}
mono_stack.clear();
for (int i = n - 1; i >= 0; --i) {
while (!mono_stack.isEmpty() && heights[mono_stack.peek()] >= heights[i]) {
mono_stack.pop();
}
//最近的小于其的右边的位置
right[i] = (mono_stack.isEmpty() ? n : mono_stack.peek());
mono_stack.push(i);
}
int ans = 0;
//固定其高度，加上拥有了左右节点
for (int i = 0; i < n; ++i) {
ans = Math.max(ans, (right[i] - left[i] - 1) * heights[i]);
}
return ans;
}
}