<leetcode>496. Next Greater Element I

496. Next Greater Element I

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].Output: [-1,3,-1]Explanation:    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.    For number 1 in the first array, the next greater number for it in the second array is 3.    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].Output: [3,-1]Explanation:    For number 2 in the first array, the next greater number for it in the second array is 3.    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

思路：

应该是最一般的解法了，遍历nums1和nuns2，在nums2中找到相同的元素的位置，再向后遍历，找是否有大于nums1中元素的元素，有就把该值赋给nums1，没有就置为-1

class Solution {    public int[] nextGreaterElement(int[] nums1, int[] nums2) {        for (int i = 0; i < nums1.length; i++) {            int temp = nums1[i];            int flag = 0;            for (int j = 0; j < nums2.length; j++) {                if(temp == nums2[j]) {                    for (int k = j; k < nums2.length; k++) {                        if(nums2[k]>temp&&flag == 0){                            flag = 1;                            nums1[i] = nums2[k];                            break;                        }else nums1[i] = -1;                    }                }            }        }        return nums1;    }}