496. 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.
Note:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

思路:为nums2建立一个map < int, int> ,key为num2各元素的值,value为元素下标.
然后遍历nums1,对其中每个元素nums1[i],找到其在nums2对应元素j,然后从j向后搜索,找到第一个大于j的就记录,为节省空间,可以直接写入nums1[i] (反正一次遍历,以后也不会再用到).
遍历结束后nums1就是输出.

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        unordered_map<int, int> hash;
        for (int i = 0; i < nums.size(); i++) {
            hash[nums[i]] = i;//建立hash映射
        }
        for (int i = 0; i < findNums.size(); i++) {
            int index = hash[findNums[i]];
            int j;
            for (j = index+1; j < nums.size(); j++) {
                if (nums[j] > nums[index]) {
                    findNums[i] = nums[j];
                    break;//找到就立即跳出本层循环
                }                
            }
            if (j >= nums.size()) findNums[i] = -1;//若没找到,则为-1
        }
        return findNums;
    }
};