LeetCode——503. Next Greater Element II

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503. Next Greater Element II

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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.

题意：

     给定一个循环数组，问按顺时针方向，比每个元素大的第一个元素分别是什么，无则为-1。

解题：

    此题解法类似于496的解法，巧妙利用栈的结构，但不同的是，保存的是还未找到比其大的元素的位置，同时又是循环数组，所以要扫两遍，第一遍扫，加入还没找到比其大的元素的位置，第二遍扫时，则无需再加入元素位置，只是给第一遍没找到的元素找对应的值，若仍无法找到，则是无解。最后再扫一遍位置，输出值即可。

代码：

class Solution {public:    vector<int> nextGreaterElements(vector<int>& nums) {        map <int,int> m;        stack <int> s;        vector <int> v;        for(int i=0;i<nums.size();i++){            if(s.empty())            {                s.push(i);            }                else{                while(!s.empty()){                    if(nums[i]>nums[s.top()]){                        m[s.top()]=nums[i];                        s.pop();                    }                    else                        break;                }                s.push(i);            }        }        for(int i=0;i<nums.size();i++){            while(!s.empty()){                if(nums[i]>nums[s.top()]){                        m[s.top()]=nums[i];                        s.pop();                    }                else break;            }            if(s.empty())break;        }        for(int i=0;i<nums.size();i++){            if(m.count(i)==0)                v.push_back(-1);            else                v.push_back(m[i]);        }        return v;    }};