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.
class Solution {
public int largestRectangleArea(int[] h) {
int n = h.length, i = 0, max = 0;
Stack<Integer> s = new Stack<>();
while (i < n) {
// as long as the current bar is shorter than the last one in the stack
// we keep popping out the stack and calculate the area based on
// the popped bar
while (!s.isEmpty() && h[i] < h[s.peek()]) {
// tricky part is how to handle the index of the left bound
max = Math.max(max, h[s.pop()] * (i - (s.isEmpty() ? 0 : s.peek() + 1)));
}
// put current bar's index to the stack
s.push(i++);
}
// finally pop out any bar left in the stack and calculate the area based on it
while (!s.isEmpty()) {
max = Math.max(max, h[s.pop()] * (n - (s.isEmpty() ? 0 : s.peek() + 1)));
}
return max;
}
}class Solution {
public int largestRectangleArea(int[] heights) {
int ret = 0;
Stack<Integer> stk = new Stack<>();
for(int i = 0; i < heights.length; i ++)
{
if(stk.empty() || stk.peek() <= heights[i])
stk.push(heights[i]);
else
{
int count = 0;
while(!stk.isEmpty() && stk.peek() > heights[i])
{
count ++;
ret = Math.max(ret, stk.peek()*count);
stk.pop();
}
while((count --)>0)
stk.push(heights[i]);
stk.push(heights[i]);
}
}
int count = 1;
while(!stk.isEmpty())
{
ret = Math.max(ret, stk.peek()*count);
stk.pop();
count ++;
}
return ret;
}
}首先想到一种最直观的方法：把平面看做大矩阵，条形看做1，非条形看做0，寻找最大全1矩阵。
思路上是正确的，可以用DP来做，不过会超时。
网上看到一种借助栈的做法，代码很漂亮，但是解释都非常模糊，我看懂之后，决定仔细描述思路如下：
1、如果已知height数组是升序的，应该怎么做？
比如1,2,5,7,8
那么就是(1*5) vs. (2*4) vs. (5*3) vs. (7*2) vs. (8*1)
也就是max(height[i]*(size-i))
2、使用栈的目的就是构造这样的升序序列，按照以上方法求解。
但是height本身不一定是升序的，应该怎样构建栈？
比如2,1,5,6,2,3
（1）2进栈。s={2}, result = 0
（2）1比2小，不满足升序条件，因此将2弹出，并记录当前结果为2*1=2。
将2替换为1重新进栈。s={1,1}, result = 2
（3）5比1大，满足升序条件，进栈。s={1,1,5},result = 2
（4）6比5大，满足升序条件，进栈。s={1,1,5,6},result = 2
（5）2比6小，不满足升序条件，因此将6弹出，并记录当前结果为6*1=6。s={1,1,5},result = 6
2比5小，不满足升序条件，因此将5弹出，并记录当前结果为5*2=10（因为已经弹出的5,6是升序的）。s={1,1},result = 10
2比1大，将弹出的5,6替换为2重新进栈。s={1,1,2,2,2},result = 10
（6）3比2大，满足升序条件，进栈。s={1,1,2,2,2,3},result = 10
栈构建完成，满足升序条件，因此按照升序处理办法得到上述的max(height[i]*(size-i))=max{3*1, 2*2, 2*3, 2*4, 1*5, 1*6}=8<10
综上所述，result=10
10/09/2019
局部峰值：当前值大于右侧
class Solution {
public int largestRectangleArea(int[] heights) {
int res = 0;
int he = heights.length;
for(int i = 0; i < he; i++){
if(i+1 < he && heights[i+1] >= heights[i]) continue;
int mh = heights[i];
for(int j = i; j >= 0; j--){
mh = Math.min(mh, heights[j]);
int area = mh * (i - j + 1);
res = Math.max(res, area);
}
}
return res;
}
}
