496. Next Greater Element I

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本文介绍了一个算法问题，即如何找到一个数组中每个元素在另一个数组中的下一个更大的元素。通过使用哈希映射记录元素位置，并遍历查找的方式解决此问题。

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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:

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

先遍历一遍nums，用map存下各个数的位置，再遍历findNums，根据map存的index向后找比他大的第一个数。如果没有，res的相应位存-1。代码如下：

public class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i ++) {
            map.put(nums[i], i);
        }
        int[] res = new int[findNums.length];
        for (int i = 0; i < findNums.length; i ++) {
            int index = map.get(findNums[i]);
            int j = 0;
            for (j = index + 1; j < nums.length; j ++) {
                if (nums[j] > nums[index]) {
                    res[i] = nums[j];
                    break;
                }
            }
            if (j == nums.length) {
                res[i] = -1;
            }
        }
        return res;
    }
}