这道题是说给了一个直方图,然后需要在里面找出一个最大的矩形。
这道题的解题方式,就是要使用一个栈,保持一个持续上升的队列
具体的请参照这个:http://blog.csdn.net/doc_sgl/article/details/11805519
我是参照这个的
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.
import java.util.*;
public class Solution {
/** * http://blog.csdn.net/doc_sgl/article/details/11805519 * * 这道题关键是要连续上升的形状,还有注意len的计算位置 * */
public int largestRectangleArea(int[] heights) {
Stack<Integer> stack = new Stack<Integer>();
int max_area=0,currentHeight,lastIndex,len;
for(int i=0;i<heights.length;i++){
currentHeight=heights[i];
//保持升序
if(stack.isEmpty() || heights[stack.peek()]<=currentHeight){
stack.push(i);
continue;
}
//到了这里,就是不满足升序,需要弹出,注意当前len的判断方式
while(stack.isEmpty()==false && heights[stack.peek()]>currentHeight){
lastIndex=stack.pop();
len=stack.isEmpty()?i:(i-stack.peek()-1);
max_area=Math.max(max_area,len*heights[lastIndex]);
}
stack.push(i);
}
//最后需要多处理一次
while(stack.isEmpty()==false){
lastIndex=stack.pop();
len=stack.isEmpty()?heights.length:(heights.length-stack.peek()-1);
max_area=Math.max(max_area,len*heights[lastIndex]);
}
return max_area;
}
}