Leetcode之 Largest Rectangle in Histogram

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.

Example:

Input: [2,1,5,6,2,3]
Output: 10

代码：

1——递归方法，较慢

class Solution {
public:
int calmin(vector<int>& height, int i, int j) {
	int index = -1;
	int min = INT_MAX;
	for (int m = i; m <= j; m++) {
		if (height[m] < min) {
			min = height[m];
			index = m;
		}
	}
	return index;
}
int helper(vector<int>& height, int i, int j) {
	if (i > j)return 0;
	if (i == j)return height[i];
	int index = calmin(height, i, j);
	int area = (j - i + 1)*height[index];
	int leftarea = helper(height, i, index - 1);
	int rightarea = helper(height, index + 1, j);
	int res = max({ area,leftarea,rightarea });
	return res;
}
int largestRectangleArea(vector<int>& heights) {
	int i = 0, j = heights.size() - 1;
	return helper(heights, i, j);
}
};

2——遍历，所有面积遍历一遍然后存起来

class Solution {
public:

int largestRectangleArea(vector<int>& heights) {
        int res = 0;
        for (int i = 0; i < heights.size(); ++i) {
            if (i + 1 < heights.size() && heights[i] <= heights[i + 1]) {
                continue;
            }
            int minH = heights[i];
            for (int j = i; j >= 0; --j) {
                minH = min(minH, heights[j]);
                int area = minH * (i - j + 1);
                res = max(res, area);
            }
        }
        return res;
}
};

想法：

多动脑子