Leetcode刷题记—— 84. Largest Rectangle in Histogram(柱形图中最大矩形面积)

一、题目叙述：

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.

For example,
Given heights = [2,1,5,6,2,3],
return 10.

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二、解题思路：

分两种情况。1：当栈空或者当前高度大于栈顶下标所指示的高度时，当前下标入栈。否则，2：当前栈顶出栈，并且用这个下标所指示的高度计算面积。而这个方法为什么只需要一个栈呢？因为当第二种情况时，for循环的循环下标回退，也就让下一次for循环比较当前高度与新的栈顶下标所指示的高度，注意此时的栈顶已经改变由于之前的出栈。

三、源码：

1、参考：

import java.util.LinkedList;public class Solution {public int largestRectangleArea(int[] height) {  // Start typing your Java solution below  // DO NOT write main() function  int area = 0;  java.util.Stack<Integer> stack = new java.util.Stack<Integer>();  for (int i = 0; i < height.length; i++) {    if (stack.empty() || height[stack.peek()] < height[i]) {      stack.push(i);    } else {      int start = stack.pop();      int width = stack.empty() ? i : i - stack.peek() - 1;      area = Math.max(area, height[start] * width);      i--;    }  }  while (!stack.empty()) {    int start = stack.pop();    int width = stack.empty() ? height.length : height.length - stack.peek() - 1;    area = Math.max(area, height[start] * width);        }  return area;}    public static void main(String args[])      {                  //  int[] digits = {0};          Solution solution = new Solution();          int[] abc = {2,1,5,6,2,3};         // for(int i = 0; i < abc.length; i ++)          System.out.print(solution.largestRectangleArea(abc));            }  }

2、超时：

import java.util.LinkedList;public class Solution {    public int largestRectangleArea(int[] heights)     {    int result = 0;        LinkedList<Integer> stack = new LinkedList<Integer>();        for (int i = 0; i < heights.length; i++)        {        if (stack.isEmpty() || stack.getLast() <= heights[i])        stack.add(heights[i]);        else        {        int count = 0;        while (!stack.isEmpty() && stack.getLast() > heights[i])        {        count ++;        //if (stack.getLast() * count > result)                result = Math.max(result, stack.getLast() * count);        //result = ;        stack.removeLast();        }        for (int j = 0; j < count; j++)        stack.add(heights[i]);        stack.add(heights[i]);        }                }        for (int i = 0; i < heights.length; i ++)        {        result = Math.max(result, stack.get(i) * (heights.length - i));        }        return result;    }    public static void main(String args[])      {                  //  int[] digits = {0};          Solution solution = new Solution();          int[] abc = {2,1,5,6,2,3};         // for(int i = 0; i < abc.length; i ++)          System.out.print(solution.largestRectangleArea(abc));            }  }