LeetCode之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.
      
     
   
  

给你两个数组，为数组1和数组2，数组1为数组2的子集。找出数组1的每一个元素在数组2中对应的元素a,然后找到元素a后侧第一个比a大的数构成一个数组，即是我们需要的答案。如果不存在，则为-1。

   

    

     

      

     
     

      

       Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
      
     
    

     

      

     
     

      

       Output: [-1,3,-1]
      
     
   
  

   

    

     

      

     
     

      

       public class Solution {
      
     
    

     

      

     
     

      

        public int[] nextGreaterElement(int[] findNums, int[] nums) {
      
     
    

     

      

     
     

      

        if (findNums == null || nums == null) {
      
     
    

     

      

     
     

      

        return null;
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        int findLength = findNums.length;
      
     
    

     

      

     
     

      

        int numsLength = nums.length;
      
     
    

     

      

     
     

      

        int[] result = new int[findLength];
      
     
    

     

      

     
     

      

        boolean flag = false;
      
     
    

     

      

     
     

      

        for (int i = 0; i < findLength; ++i) {
      
     
    

     

      

     
     

      

        for (int j = 0; j < numsLength; ++j) {
      
     
    

     

      

     
     

      

        if (findNums[i] == nums[j]) {
      
     
    

     

      

     
     

      

        if (j + 1 == numsLength) {
      
     
    

     

      

     
     

      

        result[i] = -1;
      
     
    

     

      

     
     

      

        } else {
      
     
    

     

      

     
     

      

        for (int k = j + 1; k < numsLength; ++k) {
      
     
    

     

      

     
     

      

        if (nums[j] < nums[k]) { 
      
     
    

     

      

     
     

      

        result[i] = nums[k];
      
     
    

     

      

     
     

      

        flag = true;
      
     
    

     

      

     
     

      

        break;
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        if (!flag) {
      
     
    

     

      

     
     

      

        result[i] = -1;
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        flag = false;
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        return result; 
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

       }
      
     
   
  

   

    

     

      

     
     

      

        for (int k = j + 1;  k < numsLength; ++k) {
      
     
    

     

      

     
     

      

        if (nums[j] < nums[k]) { 
      
     
    

     

      

     
     

      

        result[i] = nums[k];
      
     
    

     

      

     
     

      

        flag = true;
      
     
    

     

      

     
     

      

        break;
      
     
    

     

      

     
     

      

        }
      
     
    

     

      

     
     

      

        }