LeetCode 84. Largest Rectangle in Histogram

Problem Description:

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.

解题思路：

面积的计算肯定是长乘宽，就是index的差加一乘上最小的height。

此题O(n)的解法是维持一个递增的栈，每次出现一次递减的时候开始弹出栈中元素，维持递增性质。

最后一次的高度为0，是为了弹出所有的元素。

class Solution {
    public int largestRectangleArea(int[] heights) {
        if(heights == null) return 0;
        Stack<Integer> stack = new Stack<>();
        int res = 0;
        for(int i = 0; i <= heights.length; i++) {
            int h = i == heights.length ? 0 : heights[i];
            if(stack.isEmpty() || h >= heights[stack.peek()]) {
                stack.push(i);
            } else {
                int top = stack.pop();
                res = Math.max(res, heights[top] * (stack.isEmpty() ? i : i - 1 - stack.peek()));
                i--;
            }
        }
        return res;
    }
}

解法二是O(n^2)的复杂度。但是没有用到栈，提交后速度也很快，思路类似，每当出现height[i] > height[i + 1]的时候进行一次面积的计算。

class Solution {
    public int largestRectangleArea(int[] heights) {
        if(heights == null) return 0;
        int res = 0;
        for(int i = 0; i < heights.length; i++) {
            if(i == heights.length - 1 || heights[i] > heights[i + 1]) {
                int cur = heights[i];
                int minH = cur;
                res = Math.max(res, cur);
                for(int j = i - 1; j >= 0; j--) {
                    minH = Math.min(heights[j], minH);
                    res = Math.max(res, minH * (i - j + 1));
                }
            }
        }
        return res;
    }
}