[leetcode]503. Next Greater Element II

[leetcode]503. Next Greater Element II

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Analysis

电脑坏了—— [心塞！！]

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.

Implement

方法一：（暴力）

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int len = nums.size();
        vector<int> res(len, -1);
        for(int i=0; i<len; i++){
            for(int j=i+1; j<len+i; j++){
                if(nums[j%len] > nums[i]){
                    res[i] = nums[j%len];
                    break;
                }
            }
        }
        return res;
    }
};

方法二：（stack）

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int len = nums.size();
        vector<int> res(len, -1);
        stack<int> nums1;
        for(int i=0; i<2*len; i++){
            int num = nums[i%len];
            while(!nums1.empty() && nums[nums1.top()] < num){
                res[nums1.top()] = num;
                nums1.pop();
            }
            if(i<len)
                nums1.push(i);
        }
        return res;
    }
};