496 Next Greater Element I Next Greater Element I

Given two arrays nums1 and nums2 without repeating elements, where nums1 is a subset of nums2. Find the next greater value in nums2 for each element in nums1.
The next greater element of the number x in nums1 is the first element greater than x to the right of the corresponding position of x in nums2. If it does not exist, the corresponding position outputs -1.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For the number 4 in num1, you cannot The next higher number is found in the second array, so -1 is output.
    For number 1 in num1, the next larger number to the right of number 1 in the second array is 3.
    For number 2 in num1, there is no next larger number in the second array, so -1 is output.

Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in num1, the next in the second array The larger number is 3.
    For the number 4 in num1, there is no next larger number in the second array, so -1 is output.
Note:
    1. All elements in nums1 and nums2 are unique.
    2. The array size of nums1 and nums2 is not more than 1000.
See: https://leetcode.com/problems/next-greater-element-i/description/

C++:

method one:

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
    {
        vector<int> res(findNums.size());
        for (int i = 0; i < findNums.size(); ++i)
        {
            int j = 0, k = 0;
            for (; j < nums.size(); ++j)
            {
                if (nums[j] == findNums[i])
                {
                    break;
                }
            }
            for (k = j + 1; k < nums.size(); ++k)
            {
                if (nums[k] > nums[j])
                {
                    res[i] = nums[k];
                    break;
                }
            }
            if (k == nums.size())
            {
                res[i] = -1;
            }
        }
        return res;
    }
};

 Method Two:

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
    {
        vector<int> res(findNums.size());
        unordered_map<int, int> m;
        for (int i = 0; i < nums.size(); ++i)
        {
            m[nums[i]] = i;
        }
        for (int i = 0; i < findNums.size(); ++i)
        {
            res[i] = -1;
            int start = m[findNums[i]];
            for (int j = start + 1; j < nums.size(); ++j)
            {
                if (nums[j] > findNums[i])
                {
                    res[i] = nums[j];
                    break;
                }
            }
        }
        return res;
    }
};

 Reference: http://www.cnblogs.com/grandyang/p/6399855.html