leetCode 84. Largest Rectangle in Histogram

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本文介绍了一种使用栈来解决寻找直方图中最大矩形面积的问题的方法。通过迭代和栈操作，该算法能在O(n)的时间复杂度内找到给定直方图的最大矩形面积。

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.

返回直方图最大矩形

本来用动态规划做，但是超时了

public static int largestRectangleArea(int[] heights) {
         if(heights==null||heights.length<1)return 0;
         int n=heights.length;
         int dp[][]=new int [n+1][n+1];
         int min[][]=new int [n+1][n+1];

         int max=0;
         for(int i=1;i<=n;i++){
            min[i][i]=heights[i-1];
            dp[i][i]=heights[i-1];
            if(heights[i-1]>max)
                max=heights[i-1];
         }
         for(int i=1;i<=n;i++){
             for(int j=i+1;j<=n;j++){
                min[i][j]=heights[j-1]<min[i][j-1]?heights[j-1]:min[i][j-1];
             }
         }
         for(int i=1;i<=n;i++){
             for(int j=i+1;j<=n;j++){
                if(dp[i][j-1]>max){
                    max=dp[i][j-1];
                }

                if(min[i][j-1]>=heights[j-1]){
                    int temp=heights[j-1]*(j-i+1);

                    dp[i][j]=temp;
                    if(temp>max){
                        max=heights[j-1]*(j-i+1);

                    }
                }
                else{
                    int temp=min[i][j-1]*(j-i+1);
                    if(temp>dp[i][j]){
                        dp[i][j]=temp;
                        if(temp>max)
                        max=temp;

                    }

                }

             }
         }
        return max;

        }

能通过的是参考网上的用栈的方法解决的，思路看这里

 public static int largestRectangleArea(int[] heights) {
         if(heights==null||heights.length<1)return 0;

         int n=heights.length;
         if(n==1)return heights[0];
         int max=0;
         Stack<Integer> st = new Stack<Integer>();
         st.push(heights[0]);
         int i=1;
         while(i<n){
             if(heights[i]>=st.peek()){
                 st.push(heights[i]);
             }
             else{
                 int count1=1;
                 while(!st.isEmpty()&&st.peek()>heights[i]){
                     int temp=st.pop();
                     if(count1*temp>max){
                         max=count1*temp;
                     }
                     count1++;
                 }
                 while(count1-->0){
                     st.push(heights[i]);
                 }
             }

             i++;
         }
         int count1=1;
         while(!st.isEmpty()){
             int temp=st.pop();
             if(count1*temp>max){
                 max=count1*temp;
             }
             count1++;
         }
         return max;
        }