[LeetCode] Next Greater Element II 下一個較大的元素之二

 

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.

 

這道題是之前那道Next Greater Element I的拓展，不同的是，此時數組是一個循環數組，就是說某一個元素的下一個較大值可以在其前面，那么對於循環數組的遍歷，為了使下標不超過數組的長度，我們需要對n取余，下面先來看暴力破解的方法，遍歷每一個數字，然后對於每一個遍歷到的數字，遍歷所有其他數字，注意不是遍歷到數組末尾，而是通過循環數組遍歷其前一個數字，遇到較大值則存入結果res中，並break，再進行下一個數字的遍歷，參見代碼如下： 

 

解法一：

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

 

我們可以使用棧來進行優化上面的算法，我們遍歷兩倍的數組，然后還是坐標i對n取余，取出數字，如果此時棧不為空，且棧頂元素小於當前數字，說明當前數字就是棧頂元素的右邊第一個較大數，那么建立二者的映射，並且去除當前棧頂元素，最后如果i小於n，則把i壓入棧。因為res的長度必須是n，超過n的部分我們只是為了給之前棧中的數字找較大值，所以不能壓入棧，參見代碼如下：

 

解法二：

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

 

類似題目：

Next Greater Element I 

Next Greater Element III

 

參考資料：

https://discuss.leetcode.com/topic/77871/short-ac-solution-and-fast-dp-solution-45ms

https://discuss.leetcode.com/topic/77923/java-10-lines-and-c-12-lines-linear-time-complexity-o-n-with-explanation

 

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