LeetCode-84 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 height = [2,1,5,6,2,3]
,
return 10
.
思路:计算以height[i]为最低点时的最大面积。
1. 创建一个栈stack;
1. 从头开始遍历数组,如果数组为空或者当前遍历元素height[i]>=stack.peek(),就将"i"进栈,注意是将"i"而不是height[i];
3. 如果height[i]<stack.peek();
计算以height[stack.pop()]为最短高度,长度为i-stack.peek()-1的面积;
如果此时stack为空,则长度为"i";
如果计算得到的面积大,则更新当前最大面积max;
4. 当遍历到数组结束时,添加一个元素“0”作为终点判断。
代码如下:
1 public int largestRectangleArea(int[] height) { 2 if(height == null || height.length == 0) return 0; 3 Stack<Integer> stack = new Stack<Integer>(); 4 stack.push(-1); 5 int maxArea = 0; 6 for(int i=0; i<=height.length;) { 7 int curr = (i==height.length ? 0 : height[i]); 8 if(stack.size() == 1 || curr > height[stack.peek()]) { 9 stack.push(i); 10 i++; 11 } else { 12 int index = stack.pop(); 13 maxArea = Math.max((i-stack.peek()-1)*height[index], maxArea); 14 } 15 } 16 return maxArea; 17 }
posted on 2015-03-17 23:16 linxiong1991 阅读(121) 评论(0) 编辑 收藏 举报