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;i0 && 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

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 stack = new java.util.Stack();  
  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;  
}  
}