[LeetCode] 503. Next Greater Element II

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.

下一个更大的元素II。

版本一传送门在此。这个题依然是找数组中往右比当前元素更大的元素，但是版本二改了一个设定，需要假设input是一个环，也就是说数组最后一个元素在res里面的结果很有可能不是-1。

思路依然是单调栈，去看栈顶元素是否比当前元素小，若栈顶元素小则弹出栈并存到结果集res里面；若栈顶元素大则还是把当前元素的index入栈。单调栈类型的题绝大部分放入栈的都是遍历数组的index。这道题因为是环的关系所以遍历两遍input数组即可。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Arrays.fill(res, -1);
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < len * 2; i++) {
            int cur = nums[i % len];
            while (!stack.isEmpty() && nums[stack.peek()] < cur) {
                res[stack.pop()] = cur;
            }
            // when loop for the first time
            if (i < len)
                stack.push(i);
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var nextGreaterElements = function (nums) {
    let len = nums.length;
    let res = new Array(len).fill(-1);
    let stack = [];
    for (let i = 0; i < nums.length * 2; i++) {
        let cur = nums[i % len];
        while (stack.length > 0 && nums[stack[stack.length - 1]] < cur) {
            res[stack.pop()] = cur;
        }
        if (i < len) {
            stack.push(i);
        }
    }
    return res;
};

LeetCode 题目总结