Leetcode题解14 84. Largest Rectangle in Histogram(hard)
2016-05-06 11:11
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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.
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
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.
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
public class Solution { public static int largestRectangleArea(int[] heights) { Stack<Integer> mStack = new Stack<Integer>(); int count; int result = 0; for (int i = 0; i < heights.length; i++) { if (mStack.isEmpty() || mStack.peek() <= heights[i]) { mStack.push(heights[i]); } else { count = 1; while (!mStack.isEmpty() && mStack.peek() > heights[i]) { result = max(result, mStack.peek() * count); mStack.pop(); count++; } while (count > 1) { mStack.push(heights[i]); count--; } mStack.push(heights[i]); } } int num = 1; while (!mStack.isEmpty()) { result = max(result, mStack.pop() * num); num++; } return result; } public static int max(int i, int j) { return i >= j ? i : j; } }
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