Largest Rectangle in Histogram

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.

这个题暴力解法肯定是超时。

化繁为简
先假设后面的都比前面的大，如下图，那就容易计算了。

但是如果碰到前面比后面大的情况，就可以先将和前面比它大的全部都比较一遍，然后放入栈中，直到栈里的所有元素都是递增的。

代码如下：

public int largestRectangleArea(int[] height) {      int area = 0;      java.util.Stack<Integer> heightStack = new java.util.Stack<Integer>();      java.util.Stack<Integer> indexStack = new java.util.Stack<Integer>();      for (int i = 0; i < height.length; i++) {          if (heightStack.empty() || heightStack.peek() <= height[i]) {              heightStack.push(height[i]);              indexStack.push(i);          }        else if(heightStack.peek() > height[i]) {             int j = 0;              while (!heightStack.empty() && heightStack.peek() > height[i]) {                  j = indexStack.pop();                  int currArea = (i - j) * heightStack.pop();                  if (currArea > area) {                      area = currArea;                  }              }              heightStack.push(height[i]);              indexStack.push(j);          }       }     while (!heightStack.empty()) {           int currArea = (height.length - indexStack.pop()) * heightStack.pop();           if (currArea > area) {               area = currArea;           }       }       return area;      } 

如果我们遇到这道题的时候，一开始应该想想特例，比如递增序列下的最大矩形面积，然后发散开来，想想一般情况和这种递增情况的关系，也许就能有突破。使用类似的”从特例到一般”的发散方式还有Candy (分糖果)的第二种解法。