84直方图最大矩形覆盖 · Largest Rectangle in Histogram

［抄题］：

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

［思维问题］：

［一句话思路］：

小高度使得全部大高度都终结了，对POP出来的面积打擂台

［输入量］：空：正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

stack.peek() == 0 时属于非空
max要初始化为0
可以用height[stack.peek()]来运用数组
i = 0; i <= height.length 都要打印比如1 1

［二刷］：

用三元运算符，curt到头了应该是-1，width在栈为空时就是i本身
在括号里pop了也算是完成了pop，不用再pop了

［总结］：

［复杂度］：Time complexity: O(n) push了n个但是只会pop出来一个 Space complexity: O(n)

［英文数据结构，为什么不用别的数据结构］：

［其他解法］：

［Follow Up］：

85 1拼出的最大矩形

［题目变变变］：

public class Solution {public int largestRectangleArea(int[] height) {if (height == null || height.length == 0) {return 0;}Stack<Integer> stack = new Stack<Integer>();int max = 0;for (int i = 0; i <= height.length; i++) {int curt = (i == height.length) ? -1 : height[i];while (!stack.isEmpty() && curt <= height[stack.peek()]) {int h = height[stack.pop()];int w = stack.isEmpty() ? i : i - stack.peek() - 1;max = Math.max(max, h * w);}stack.push(i);}return max;}
}

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转载于:https://www.cnblogs.com/immiao0319/p/8250154.html

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