[Monotone Stack] LeetCode-503. The Next Bigger Element II

503. The Next Greater Element II

Title description

Given a looping array (the next element of the last element is the first element of the array), output the next larger element of each element. The next larger element of the number x is in the traversal order of the array. The first number after this number is larger than it, which means you should search for the next larger number in a loop. If it does not exist, -1 is output.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The next greater number of the first 1 is 2; the
number 2 cannot find the next greater number; the
second The next largest number of 1 requires a loop search, and the result is also 2.

Problem-solving ideas

Monotonic stack

    public static int[] nextGreaterElements(int[] nums) {
    
    
        int length = nums.length;
        int res[] = new int[length];
        Arrays.fill(res,-1); //默认都为1
        Stack<Integer> stack = new Stack<>();
        //相当于把数组遍历两遍
        for (int i = 0; i < length * 2; i++) {
    
    
            //遍历数组的第index（index从0开始）个元素，因为数组会遍历
            //两遍，会导致数组越界异常，所以这里要对数组长度进行求余
            int index = i % length;
            //单调栈，他存储的是元素的下标，不是元素具体值，从栈顶
            //到栈底所对应的值是递增的（栈顶元素在数组中对应的值最小，
            //栈底元素对应的值最大），如果栈顶元素对应的值比nums[index]小，
            //说明栈顶元素对应的值遇到了右边第一个比他大的值，然后栈顶元素出栈，
            //让他对应的位置变为nums[index]，也就是他右边第一个比他大的值，
            //然后继续判断……
            while (!stack.isEmpty() && nums[stack.peek()] < nums[index]){
    
    
                res[stack.pop()] = nums[index];
            }
            //当前元素的下标入栈
            stack.push(index);
        }
        return res;
    }