[496]下一个更大元素 I

参考：
https://leetcode-cn.com/problems/next-greater-element-i/solution/xia-yi-ge-geng-da-yuan-su-i-by-leetcode/

package T496下一个更大元素I;

import java.util.HashMap;

public class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            map.put(nums2[i], i);
        }
        int[] result = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            boolean flag = false;
            int start = map.get(nums1[i]);
            for (int j = start+1; j < nums2.length; j++) {
                if (nums2[j] > nums1[i]) {
                    result[i] = nums2[j];
                    flag = true;
                    break;
                }
            }
            if (!flag) {
                result[i] = -1;
            }
        }
        return result;
    }
}

```
class Solution:
    def nextGreaterElement(self, nums1, nums2):
        m = dict()
        nums2_len = len(nums2)
        for index, num in enumerate(nums2):
            m[num] = index
        result = list()
        for i in nums1:
            index = m[i]
            # 是否找到更大的值
            flag = False
            for next_num_index in range(index,nums2_len):
                if nums2[next_num_index] > i:
                    result.append(nums2[next_num_index])
                    flag = True
                    break
            if flag is False:
                result.append(-1)
        return result


if __name__ == '__main__':
    x = Solution()
    nums1 = [4, 1, 2]
    nums2 = [1, 3, 4, 2]
    print(x.nextGreaterElement(nums1, nums2))