LeetCode 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.

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

Runtime: 3 ms runtime beats 99.39% of java submissions.

	public int largestRectangleArea(int[] heights) {
		if (heights == null || heights.length == 0) return 0;
		return getMax(heights, 0, heights.length);
	}

	int getMax(int[] heights, int s, int e) {
		if (s + 1 >= e) return heights[s];
		int min = s;//柱状图的最低点
		boolean sorted = true;//所有柱子是否按照升序
		for (int i = s; i < e; i++) {
			if (i > s && heights[i] < heights[i - 1]) sorted = false;
			if (heights[min] > heights[i]) min = i;
		}
		if (sorted) {//如果正序的话就简单了直接计算
			int max = 0;
			for (int i = s; i < e; i++) {
				max = Math.max(max, heights[i] * (e - i));
			}
			return max;
		}
		//不是正序的话,就以最低点为准,划分左右两个区域分别计算,最后再和加上min的比较
		int left = (min > s) ? getMax(heights, s, min) : 0;
		int right = (min < e - 1) ? getMax(heights, min + 1, e) : 0;
		return Math.max(Math.max(left, right), (e - s) * heights[min]);
	}

使用栈。参考http://www.cnblogs.com/yrbbest/p/4437139.html Runtime: 21 ms

	public int largestRectangleArea2(int[] height) {
		if (height == null || height.length == 0) return 0;
		Stack<Integer> stack = new Stack<Integer>();
		int max = 0;
		for (int i = 0; i <= height.length; i++) {
			int curt = (i == height.length) ? -1 : height[i];
			while (!stack.isEmpty() && curt <= height[stack.peek()]) {
				int h = height[stack.pop()];
				int w = stack.isEmpty() ? i : i - stack.peek() - 1;
				max = Math.max(max, h * w);
			}
			stack.push(i);
		}

		return max;
	}