下一个更大元素II

503. 下一个更大元素 II - 力扣（LeetCode）

最直接的方法，我自己写的。。

class Solution {public int[] nextGreaterElements(int[] nums) {int len = nums.length*2;int [] nums2 = new int[len];for(int i = 0;i<len;i++){if(i<len/2){nums2[i]=nums[i];}else{nums2[i]=nums[i-len/2];}}int[] res = new int[len/2];Arrays.fill(res,-1);Deque<Integer> statk = new LinkedList<>();for(int i = 0;i<len;i++){while(!statk.isEmpty()&&nums2[i]>nums2[statk.peek()]){int pre = statk.pop();if(pre<len/2){res[pre] = nums2[i];}}statk.push(i);}return res;}
}

优化

class Solution {public int[] nextGreaterElements(int[] nums) {//边界判断if(nums == null || nums.length <= 1) {return new int[]{-1};}int size = nums.length;int [] res = new int[size];Arrays.fill(res,-1);Deque<Integer> stack = new LinkedList<>();for(int i = 0;i<2*size;i++){while(!stack.isEmpty()&&nums[i%size]>nums[stack.peek()]){int pre = stack.pop();res[pre] = nums[i%size];}stack.push(i%size);}return res;}
}

接雨水

42. 接雨水 - 力扣（LeetCode）

class Solution {public int trap(int[] height) {if(height.length<=2){return 0;}int res = 0;Deque<Integer> stack = new LinkedList<>();for(int i = 0;i<height.length;i++){while(!stack.isEmpty()&&height[i]>height[stack.peek()]){int right = height[i];int midIndex = stack.pop();int mid = height[midIndex];if(!stack.isEmpty()){int leftIndex = stack.peek();int left = height[leftIndex];int h = Math.min(right,left) - mid;int w = i-leftIndex-1;res+=w*h;}}stack.push(i);}return res;}
}