题目：
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.
题解：
这道题自己是完全没想到用栈了。。
有个很完整很详细很好的讲解在这里： http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html
我就不写了。贴一下上面提到的代码吧。
O(n^2)的：
1 public int largestRectangleArea(int[] height) {
2 // Start typing your Java solution below
3 // DO NOT write main() function
4 int[] min = new int[height.length];
5 int maxArea = 0;
6 for(int i = 0; i < height.length; i++){
7 if(height[i] != 0 && maxArea/height[i] >= (height.length - i)) {
8 continue;
9 }
10 for(int j = i; j < height.length; j++){
11 if(i == j) min[j] = height[j];
12 else {
13 if(height[j] < min[j - 1]) {
14 min[j] = height[j];
15 }else min[j] = min[j-1];
16 }
17 int tentativeArea = min[j] * (j - i + 1);
18 if(tentativeArea > maxArea) {
19 maxArea = tentativeArea;
20 }
21 }
22 }
23 return maxArea;
24 }
2 // Start typing your Java solution below
3 // DO NOT write main() function
4 int[] min = new int[height.length];
5 int maxArea = 0;
6 for(int i = 0; i < height.length; i++){
7 if(height[i] != 0 && maxArea/height[i] >= (height.length - i)) {
8 continue;
9 }
10 for(int j = i; j < height.length; j++){
11 if(i == j) min[j] = height[j];
12 else {
13 if(height[j] < min[j - 1]) {
14 min[j] = height[j];
15 }else min[j] = min[j-1];
16 }
17 int tentativeArea = min[j] * (j - i + 1);
18 if(tentativeArea > maxArea) {
19 maxArea = tentativeArea;
20 }
21 }
22 }
23 return maxArea;
24 }
O(n)的：
1 public int largestRectangleArea2(int[] height) {
2 Stack<Integer> stack = new Stack<Integer>();
3 int i = 0;
4 int maxArea = 0;
5 int[] h = new int[height.length + 1];
6 h = Arrays.copyOf(height, height.length + 1);
7 while(i < h.length){
8 if(stack.isEmpty() || h[stack.peek()] <= h[i]){
9 stack.push(i++);
10 }else {
11 int t = stack.pop();
12 maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
13 }
14 }
15 return maxArea;
16 }
2 Stack<Integer> stack = new Stack<Integer>();
3 int i = 0;
4 int maxArea = 0;
5 int[] h = new int[height.length + 1];
6 h = Arrays.copyOf(height, height.length + 1);
7 while(i < h.length){
8 if(stack.isEmpty() || h[stack.peek()] <= h[i]){
9 stack.push(i++);
10 }else {
11 int t = stack.pop();
12 maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
13 }
14 }
15 return maxArea;
16 }