LeetCode 84. Largest Rectangle in Histogram

Solution1：我的答案
循环里头套了一个动态规划，缺点是当heights个数或最大高度多高时会很耗时间！！！运行至第37个case时超时了….
还是记录一下。。。

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int n = heights.size();if (n == 0) return 0;else if (n == 1) return heights[0];int height_max = INT_MIN;set<int> res;for (int i = 0; i < heights.size(); i++) height_max = max(height_max, heights[i]);for (int i = 0; i <= height_max; i++) {int temp = 0, temp_max = INT_MIN;vector<int> dp(n, 0);for (int j = 0; j < n; j++) {//DP，思路与在0、1组成的字符串里求最长1的串思路相同if (heights[j] < i)temp = 0;else {temp += i;dp[j] = temp;}temp_max = max(temp_max, dp[j]);}res.insert(temp_max);}return *(--res.end());}
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4322653.html
每找到一个局部峰值，然后向前遍历所有的值，算出共同的矩形面积，每次对比保留最大值。
相当于只在每个峰值和下坡路上遍历，按理说复杂度是O(n2)O(n2)O(n^2)。但是确实很快而且可以AC。。。

class Solution {
public:int largestRectangleArea(vector<int>& height) {int res = 0;for (int i = 0; i < height.size(); ++i) {if (i + 1 < height.size() && height[i] <= height[i + 1]) {//若把这个判断条件去掉，则是暴力算法复杂度是O(n^2)，只有最后一个case无法ACcontinue;}int minH = height[i];for (int j = i; j >= 0; --j) {minH = min(minH, height[j]);int area = minH * (i - j + 1);res = max(res, area);}}return res;}
};

正如注释中所说，此算法和暴力的时间复杂度相同，而暴力算法也只是最后一个case超时而已，不过只在峰值和下坡路上遍历确实很适合AC最后一个case…

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