刷题84. Largest Rectangle in Histogram

一、题目说明

题目84. Largest Rectangle in Histogram，给定n个非负整数（每个柱子宽度为1）形成柱状图，求该图的最大面积。题目难度是Hard！

二、我的解答

这是一个 看起来容易，做起来很容易错的题目。我开始用的是“挖坑法”，遗憾的是总是Time Limit Exceeded。经过10次优化，还是很难看。

class Solution{
	public:
		int largestRectangleArea(vector<int>& heights){
			int len = heights.size();
			if(len<1) return 0;
			if(len==1) return heights[0];
			int result = 0;
			int start=len-1,end=0;
			int cnt = 0,sum=0;
			int min;

			while(1){
				min = INT_MAX;
				for(int i=0;i<len;i++){
					if(heights[i]>0 && heights[i]<min){
						min = heights[i];
					}
				}

				//找到第1个正的 
				while(end<len && heights[end]<=0){
					end++;
				}
				while(start>0 && heights[start]<=0){
					start--;
				}
				if(start==end){
					sum = heights[start];
					if(sum>result) result = sum;
					break;
				}else if(end>start) {
					break;
				}
					
				sum = 0;
				cnt = 0;
				//一次遍历，求大于0的连续长度  最大值 
				while(end<len){
					cnt = 0;
					sum = 0;
					while(end<len && heights[end]>=min){
						if(heights[end]==min) heights[end] = 0;
						cnt++;
						end++;
					}
					sum = cnt * min;
					if(sum>result) result = sum;
					if(end<len) end++;
				}
				end = 0;
				start = len-1;
			}
			return result;
		}
};

性能：

Runtime: 1536 ms, faster than 4.53% of C++ online submissions for Largest Rectangle in Histogram.
Memory Usage: 9.9 MB, less than 94.29% of C++ online submissions for Largest Rectangle in Histogram.

还能想到的方法就是暴力计算法，性能也差不多：

class Solution{
	public:
		//brute force
		int largestRectangleArea(vector<int>& heights){
			int len = heights.size();
			if(len<1) return 0;
			if(len==1) return heights[0];
			int result = 0;
			bool allthesame = true;
			for(int i=0;i<len-1;i++){
				if(heights[i]!=heights[i+1]){
					allthesame = false;
				}
			}
			if(allthesame){
				return heights[0]*len;
			}
			
			for(int i=0;i<len;i++){
				int minHeight = INT_MAX;
				for(int j=i;j<len;j++){
					minHeight = min(minHeight,heights[j]);
					result = max(result,minHeight * (j-i+1));
				}
			}
			
			return result;
		}
};

Runtime: 1040 ms, faster than 5.03% of C++ online submissions for Largest Rectangle in Histogram.
Memory Usage: 10 MB, less than 94.29% of C++ online submissions for Largest Rectangle in Histogram.

三、优化措施

用单调递增stack法，代码如下：

class Solution{
	public:
		//单调递增栈 
		int largestRectangleArea(vector<int>& heights){
			int result = 0;
			stack<int> st;
			st.push(-1);//-1 放进栈的顶部来表示开始

            //按照从左到右的顺序，我们不断将柱子的序号放进栈中,直到 heights[i]<heights[st.top]
            //将栈中的序号弹出，直到heights[stack[j]]≤heights[i]
			for(int i=0;i<heights.size();i++){
				while(st.top()!=-1 && heights[i]<heights[st.top()]){
					int h = st.top();
					st.pop();
					result = max(result,heights[h]*(i - st.top() -1));
				}
				st.push(i);
			} 
			
			// 遍历完了，但是没计算完
	        while(st.top() != -1){
	        	int h = st.top();
	        	st.pop();
	        	int len = heights.size() - st.top() -1;
	            result = max(result,heights[h]*len);
	        }
			
			return result;
		}
};

Runtime: 16 ms, faster than 53.51% of C++ online submissions for Largest Rectangle in Histogram.
Memory Usage: 10.4 MB, less than 91.43% of C++ online submissions for Largest Rectangle in Histogram.

继续优化：

class Solution{
	public:
		//单调递增栈 ,借用i当栈 
		int largestRectangleArea(vector<int>& heights){
			int result = 0;
	        int len, wid;
	        for (int i = 0; i < heights.size(); i++) {
	            if(i != heights.size() - 1 && heights[i] <= heights[i + 1]) continue;   //这一步的判断很玄妙
	            wid = heights[i];
	            for (int j = i; j >= 0; j--) {
	                len = i - j + 1;
	                wid = min(wid, heights[j]);
	                result = max(result, len * wid);
	            }
	        }
			
			return result;
		}
};

Runtime: 12 ms, faster than 89.13% of C++ online submissions for Largest Rectangle in Histogram.
Memory Usage: 10 MB, less than 94.29% of C++ online submissions for Largest Rectangle in Histogram.
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