[leetcode] 503. Next Greater Element II

原创

是念 2022-08-11 17:39:00 博主文章分类：leetcode ©著作权

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Description

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.

分析

题目的意思是：就是给你一个数组，然后找出每个位置上比这个位置（按顺序）上的数稍大的一个数。看例子就可以懂。

遍历两倍的数组，然后还是坐标i对n取余，取出数字，如果此时栈不为空，且栈顶元素小于当前数字，说明当前数字就是栈顶元素的右边第一个较大数，那么建立二者的映射，并且去除当前栈顶元素，最后如果i小于n，则把i压入栈。因为res的长度必须是n，超过n的部分我们只是为了给之前栈中的数字找较大值，所以不能压入栈。

代码

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int n=nums.size();
        vector<int> res(n,-1);
        stack<int> s;
        for(int i=0;i<2*n;i++){
            int num=nums[i%n];
            while(!s.empty()&&nums[s.top()]<num){
                res[s.top()]=num;
                s.pop();
            }
            if(i<n) s.push(i);
        }
        return res;
    }
};