class Solution {public int[] nextGreaterElements(int[] nums) {int len = nums.length;int[] res = new int[len];Deque<Integer> stack = new LinkedList<>();Arrays.fill(res,-1);for(int i = 0 ;i<len*2; i++){//循环数用取模的方式int j = i % len;while(!stack.isEmpty() && nums[j] > nums[stack.peek()] ){res[stack.peek()] = nums[j];stack.pop();}stack.push(j);}return res;}
}

class Solution {public int trap(int[] height) {int res = 0;if (height.length <= 2) return res;Deque<Integer> stack = new LinkedList<>();stack.push(0);for(int i = 0 ;i<height.length;i++){if(stack.isEmpty() || height[i] <= height[stack.peek()]){//如果栈为空，或当前柱子小于上一个柱子-》放入栈中stack.push(i);}while(height[i] >= height[stack.peek()]){int low = stack.pop();if (stack.isEmpty()) break;int dis =  i - stack.peek() -1;res += (Math.min(height[i],height[stack.peek()])-height[low]) * dis;}stack.push(i);}return res;}
}

比较左右大小都可以用单调栈，如果求右边比他大的，栈单调递增

class Solution {public int largestRectangleArea(int[] heights) {//数组扩容int[] h = new int[heights.length +2];h[0] = 0;h[h.length-1]=0;for(int i = 1;i<=heights.length;i++){h[i] = heights[i-1];}int res = 0;if (heights.length < 1) return res;Deque<Integer> stack = new LinkedList<>();stack.push(0);for(int i = 1 ;i<h.length;i++){if(h[i] >= h[stack.peek()]){//如果栈为空，或当前柱子大于上一个柱子-》放入栈中,栈中单增stack.push(i);}while(!stack.isEmpty() && h[i] < h[stack.peek()]){int mid = stack.pop();int left = stack.peek();int dis =  i - left -1;res = Math.max(res, h[mid] * dis);}stack.push(i);}return res;}
}