【C++】【LeetCode】496. Next Greater Element I

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.

解法：

使用栈存储还没有找到next great element的数，当下一个数A比栈顶数Z大，pop栈顶，并将相应数组m中下标为Z的位置保存为A，然后继续查看下一个栈顶。最后使得数组m中每个下标对应的值为该下标的next great element，如果该位置无值，返回-1。

class Solution {
public:
    vector nextGreaterElement(vector& findNums, vector& nums) {
        bool isRight;
        vector result;
        for (int i = 0; i < findNums.size(); i++) {
            isRight = false;
            for (int j: nums){
                if (!isRight && j == findNums[i]) {
                    isRight = true;
                    continue;
                }else if (isRight && j > findNums[i]) {
                    result.push_back(j);
                    break;
                }
            }
            if (result.size() != i+1) {
                result.push_back(-1);
            }
        }
        return result;

    }
};