LeetCode之Next Greater Element I

1、題目

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.      

給你兩個數組，為數組1和數組2，數組1為數組2的子集。找出數組1的每一個元素在數組2中對應的元素a,然後找到元素a後側第一個比a大的數構成一個數組，即是我們需要的答案。如果不存在，則為-1。

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

Output: [-1,3,-1]

2、代碼實作

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;    
    }
}      

3、總結我遇到的問題

我一開始寫這裡的代碼的時候

for (int k = j + 1;  k < numsLength; ++k) {
                            if (nums[j] < nums[k]) { 
                                result[i] = nums[k];
                                flag = true;
                                break;
                            }
                        }      

忘記了寫break;

然後導緻第一個元素符合條件了，但是後面也有符合條件的，導緻result[i]，取到的元素是最後面符合條件的元素，就錯了，是以，這個時候條件是說，得到第一個大于前面的數

就可以了，我們需要用break跳出循環，防穿透，防止繼續便利後面的元素，是以以後程式設計，看到符合後面第一個元素滿足要求的時候,要記得用break;形成條件反射。