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题目

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.

我的想法

正向走一遍，再反向走一遍把第一遍没找到的与前面比较

我觉得这道题的test cases真的傻逼透了。找不到返回-1，为什么要有[1,8,-1,-100,-1,222,1111111,-111111]这样的test case？？？？？？？真的写的火大

class Solution {
       public int[] nextGreaterElements(int[] nums) {
           if(nums == null || nums.length <= 0) {
               return new int[0];        }        if(nums.length == 1) {
               int[] res = {
   -1};            return res;        }        int[] res = new int[nums.length];        Arrays.fill(res, -1);        int idxRes = 0;        while(idxRes < nums.length - 1) {
               int idxNums = idxRes + 1;            while(idxNums < nums.length) {
                   if(nums[idxNums] > nums[idxRes]) {
                       res[idxRes] = nums[idxNums];                    break;                }                idxNums++;            }            idxRes++;        }                idxRes = 1;        while(idxRes < nums.length) {
               int idxNums = 0;            while(res[idxRes] != -1) {
                   idxRes++;                if(idxRes >= nums.length) {
                       break;                }            }            while(idxNums <= idxRes) {
                   if(nums[idxNums] > nums[idxRes]) {
                       res[idxRes] = nums[idxNums];                    break;                } else {
                       idxNums++;                }            }            idxRes++;        }        return res;    }}

解答

Leetcode solution: Stack

很聪明的方法，用一个stack来模拟循环的过程。感觉循环类的题都可以用stack来模拟

class Solution {
       public int[] nextGreaterElements(int[] nums) {
           Stack
   
     stack = new Stack<>();        int[] res = new int[nums.length];        for(int i = nums.length - 1; i >= 0; i--) {
               stack.push(nums[i]);        }        for(int i = nums.length - 1; i >= 0; i--) {
               while(!stack.isEmpty() && stack.peek() <= nums[i]) {
                   stack.pop();            }            if(stack.isEmpty()) {
                   res[i] = -1;            } else {
                   res[i] = stack.peek();            }            //把自己push进去，便于前一个元素比较            stack.push(nums[i]);        }        return res;    }}
   

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