496. Next Greater Element I \ 503. Next Greater Element II

496. Next Greater Element I

题目描述

ou 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.

代码实现

很简单就可以想到O(n^2)的算法：

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        int f_len = findNums.size();
        int n_len = nums.size();
        vector<int> res;

        for(int i = 0; i < f_len; i++) {
            bool flg = false;
            for(int j = 0; j < n_len; j++) {
                if(findNums[i] == nums[j]) flg = true;
                if(flg && nums[j] > findNums[i]) { res.push_back(nums[j]); flg = false; break; }
            }
            if(flg) res.push_back(-1);
        }

        return res;
    }
};

但是这种方法效率低下。
修改一下算法，我们如果可以找到其中的对应关系。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        int n_len = nums.size();
        vector<int> res, s(n_len);
        unordered_map<int, int> next; // map: n in nums -> next greater in nums
        for (int i=0, j=0; j < n_len; s[i++] = nums[j++])
            while (i && s[i-1] < nums[j]) next[s[--i]] = nums[j];

        for (int n:findNums) res.push_back(next.count(n)? next[n] : -1);
        return res;
    }
};

使用栈的方法：

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        stack<int> s;
        unordered_map<int, int> m;
        for (int n : nums) {
            while (s.size() && s.top() < n) {
                m[s.top()] = n;
                s.pop();
            }
            s.push(n);
        }
        vector<int> ans;
        for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1);
        return ans;
    }
};

503. Next Greater Element II

题目描述

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, output -1 for this number.

Example 1:
Input: [1,2,1]
Output: [2,-1,2]

Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.
Note: The length of given array won't exceed 10000.

代码实现

这里其实最重要的就是解决数字会重复出现的问题。这个时候不能使用map，所以我使用vector+pair的形式来实现可以有重复元素的映射。

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        stack<int> stk;
        int n_len = nums.size();
        int c_len = (n_len << 1) - 1;
        vectorint, int>> vp(n_len);
        vector<int>res;

        for(int i = 0; i < c_len; i++) {
            int rind;
            int ind = rind = i%n_len, tmp = nums[ind];
            while(!stk.empty() && stk.top() < tmp) {
                ind = (ind == 0)?n_len-1:ind-1;
                int flg = 0;
                while(vp[ind].first)  ind = (ind - 1 >= 0)?ind-1:n_len-1;
                vp[ind].second = tmp;
                vp[ind].first = 1;
                stk.pop();
            }
            if(!vp[rind].first)  stk.push(tmp);
        }

        for(auto n:vp)
            res.push_back(n.first?n.second:-1);

        return res;
    }
};