84. Largest Rectangle in Histogram(dp)

题目：

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.

解法一：

设置两个数组left,right，记录每个点能到最左和最右的坐标，时间复杂度O(n)

public class Solution {    public int largestRectangleArea(int[] heights) {        if(heights.length==0)return 0;        int[] left = new int[heights.length];        int[] right = new int[heights.length];        left[0]=0;        for(int i=1;i<heights.length;i++)        {        int k=i;        while(k>0 && heights[i]<=heights[k-1])        {        k=left[k-1];        }        left[i]=k;        }        right[heights.length-1]=heights.length-1;        for(int i=heights.length-2;i>=0;i--)        {        int k=i;        while(k<heights.length-1 && heights[i]<=heights[k+1])        {        k=right[k+1];        }        right[i]=k;        }        int max = -1;        for(int i=0;i<heights.length;i++)        {        int temp = heights[i]*(right[i]-left[i]+1);        max = Math.max(max, temp);        }        return max;    }}

解法二：

采用栈：

public class Solution {// O(n) using one stack  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;  }  }