LeetCode [栈] 496.Next Greater Element I（C++和Python实现）

496 Next Greater Element I [难度：简单] [哈希表/栈]

【题目】

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:

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

【题解C++】

先上个暴力解法吧~思路大概是这样：首先看清题意，不是同一位置右边的第一个最大值，而是同一个值右边的最大值！既然是靠值来联系两个数组，那第一个就会想到哈希表啦。先遍历大的数组，找到每个值对应的右边第一个最大值，将二者一个作为键一个作为值。最后再遍历第一个数组，push_back。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        int i,j,hash[100000];
        vector<int> out;
        //建立映射表
        for(i=0;i<nums2.size();i++)
        {
            //当前值
            int tmp = nums2[i];
            for(j=i+1;j<nums2.size();j++)
            {    
                //如果右边有比当前值大的，存入哈希表
                if(nums2[j]>tmp)
                {
                    hash[tmp] = nums2[j];
                    break;
                }          
            }
            //否则将-1存入哈希表
            if(j==nums2.size())
                hash[tmp] = -1;
                
        }
        for(i=0;i<nums1.size();i++)
            out.push_back(hash[nums1[i]]);
        return out;
    }
};

接下来优化一下~既然是栈的合集，那就用栈来实现8！

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        vector<int> out;
        stack<int> st;
        map<int,int> mp;
        //倒着遍历大数组
        for(int i=nums2.size()-1;i>=0;i--)
        {
            //当前数组值
            int now = nums2[i];
            //只要栈不为空&&栈顶值不大于当前值，pop以寻找更大的值
            while(!st.empty()&&st.top()<=now)
                st.pop();
            //如果栈为空，说明没找到比当前值更大的，赋值为-1，否则赋值为栈顶值
            mp[now] = st.empty()?-1:st.top();
            //入栈
            st.push(now);
        }
        for(int i=0;i<nums1.size();i++)
            out.push_back(mp[nums1[i]]);
        return out;
    }
};

【解题Python】

Python用的暴力，思路和C++也是一样的。

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        st = []
        for i in nums1:
            index = nums2.index(i)
            flag = 0
            for j in nums2[index:]:
                if j>i:
                    st.append(j)
                    flag = 1
                    break
            if not flag:
                st.append(-1)
        return st