Largest Rectangle in Histogram leetcode java

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：

public int largestRectangleArea(int[] height) {
// Start typing your Java solution below
// DO NOT write main() function
int[] min = new int[height.length];
int maxArea = 0;
for(int i = 0; i < height.length; i++){
if(height[i] != 0 && maxArea/height[i] >= (height.length - i)) {
continue;
}
for(int j = i; j < height.length; j++){
if(i == j) min[j] = height[j];
else {
if(height[j] < min[j - 1]) {
min[j] = height[j];
}else min[j] = min[j-1];
}
int tentativeArea = min[j] * (j - i + 1);
if(tentativeArea > maxArea) {
maxArea = tentativeArea;
}
}
}
return maxArea;
}
O(n)的： 

public int largestRectangleArea2(int[] height) {
Stack<Integer> stack = new Stack<Integer>();
int i = 0;
int maxArea = 0;
int[] h = new int[height.length + 1];
h = Arrays.copyOf(height, height.length + 1);
while(i < h.length){
if(stack.isEmpty() || h[stack.peek()] <= h[i]){
stack.push(i++);
}else {
int t = stack.pop();
maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
}
}
return maxArea;
}