496. Next Greater Element I

496. Next Greater Element I

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• Difficulty: Easy
• Contributors:love_Fawn

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.

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题意：

数组1是数组2的子集，遍历数组1中的每一个数，找到在数组2中对应的位置，然后向右寻找一个比它大的数，如果没有则返回-1.

算法思路：

遍历数组1中的每一个数，先寻找数组二中对应的位置，记录在k中，内层循环从k+1开始，返回第一个比数组1中数大的数，否则返回-1

注意的是因为数组的大小不确定， 要使用ArrayList，在转为int[]时，不能直接转化，要进行遍历赋值list.get

代码：

package easy;import java.util.ArrayList;public class NextGreaterElement1 {public static int[] nextGreaterElement(int[] findNums, int[] nums) {        ArrayList list = new ArrayList();                boolean flag = false;                            for(int i=0; i<findNums.length; i++){        flag = false;        int k = 0;        //找到第一个数组中的数字对应数组二中的位置        for(int t=0; t<nums.length; t++){        if(nums[t] == findNums[i]){        k = t;        break;        }        }        //寻找greater数        for(int j=k; j<nums.length; j++){        if(nums[j] > findNums[i]){        list.add(nums[j]);        flag = true;        break;        }        }        if(!flag){        list.add(-1);        }        }                int[] result = new int[list.size()];        for(int i=0; i<list.size(); i++){        result[i] = (int)list.get(i);        }                return result;    }public static void main(String[] args){int[] num1 = {4,1,2};int[] num2 = {1,3,4,2};int[] result = nextGreaterElement(num1, num2);for(int i=0; i<result.length; i++){System.out.println(result[i]);}}}