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;
}
};

参考文献

​[LeetCode] Next Greater Element II 下一个较大的元素之二​