84. Largest Rectangle in Histogram

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.

class Solution {
    
       public int largestRectangleArea(int[] h) {
  int n = h.length, i = 0, max = 0;
    
  Stack<Integer> s = new Stack<>();
    
  while (i < n) {
    // as long as the current bar is shorter than the last one in the stack
    // we keep popping out the stack and calculate the area based on
    // the popped bar
    while (!s.isEmpty() && h[i] < h[s.peek()]) {
      // tricky part is how to handle the index of the left bound
      max = Math.max(max, h[s.pop()] * (i - (s.isEmpty() ? 0 : s.peek() + 1)));
    }
    // put current bar's index to the stack
    s.push(i++);
  }
    
  // finally pop out any bar left in the stack and calculate the area based on it
  while (!s.isEmpty()) {
    max = Math.max(max, h[s.pop()] * (n - (s.isEmpty() ? 0 : s.peek() + 1)));
  }
    
  return max;

    }
}

参考 https://www.cnblogs.com/ganganloveu/p/4148303.html

class Solution {
    public int largestRectangleArea(int[] heights) {
                int ret = 0;
        Stack<Integer> stk = new Stack<>();
        for(int i = 0; i < heights.length; i ++)
        {
            if(stk.empty() || stk.peek() <= heights[i])
                stk.push(heights[i]);
            else
            {
                int count = 0;
                while(!stk.isEmpty() && stk.peek() > heights[i])
                {
                    count ++;
                    ret = Math.max(ret, stk.peek()*count);
                    stk.pop();
                }
                while((count --)>0)
                    stk.push(heights[i]);
                stk.push(heights[i]);
            }
        }
        int count = 1;
        while(!stk.isEmpty())
        {
            ret = Math.max(ret, stk.peek()*count);
            stk.pop();
            count ++;
        }
        return ret;
    }
}

首先想到一种最直观的方法：把平面看做大矩阵，条形看做1，非条形看做0，寻找最大全1矩阵。

思路上是正确的，可以用DP来做，不过会超时。

 

网上看到一种借助栈的做法，代码很漂亮，但是解释都非常模糊，我看懂之后，决定仔细描述思路如下：

1、如果已知height数组是升序的，应该怎么做？

比如1,2,5,7,8

那么就是(1*5) vs. (2*4) vs. (5*3) vs. (7*2) vs. (8*1)

也就是max(height[i]*(size-i))

2、使用栈的目的就是构造这样的升序序列，按照以上方法求解。

但是height本身不一定是升序的，应该怎样构建栈？

比如2,1,5,6,2,3

（1）2进栈。s={2}, result = 0

（2）1比2小，不满足升序条件，因此将2弹出，并记录当前结果为2*1=2。

将2替换为1重新进栈。s={1,1}, result = 2

（3）5比1大，满足升序条件，进栈。s={1,1,5},result = 2

（4）6比5大，满足升序条件，进栈。s={1,1,5,6},result = 2

（5）2比6小，不满足升序条件，因此将6弹出，并记录当前结果为6*1=6。s={1,1,5},result = 6

2比5小，不满足升序条件，因此将5弹出，并记录当前结果为5*2=10（因为已经弹出的5,6是升序的）。s={1,1},result = 10

2比1大，将弹出的5,6替换为2重新进栈。s={1,1,2,2,2},result = 10

（6）3比2大，满足升序条件，进栈。s={1,1,2,2,2,3},result = 10

栈构建完成，满足升序条件，因此按照升序处理办法得到上述的max(height[i]*(size-i))=max{3*1, 2*2, 2*3, 2*4, 1*5, 1*6}=8<10

综上所述，result=10

 

10/09/2019

 

 局部峰值：当前值大于右侧

class Solution {
    public int largestRectangleArea(int[] heights) {
        int res = 0;
        int he = heights.length;
        for(int i = 0; i < he; i++){
            if(i+1 < he && heights[i+1] >= heights[i]) continue;
            int mh = heights[i];
            for(int j = i; j >= 0; j--){
                mh = Math.min(mh, heights[j]);
                int area = mh * (i - j + 1);
                res = Math.max(res, area);
            }
        }
        return res;
    }
}