503. Next Greater Element II 503. Next Greater Element II

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. 503. Next Greater Element II

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是LeetCode Next Greater Element I的扩展，本题的数组可能包含重复元素，且是循环数组，即往右找到尾之后，可以接着从头开始找更大的数。 同样使用栈，只不过需要扫描两遍数组，且栈中保存的是元素下标，理解之后发现挺巧妙的。 比如nums = [9, 8, 7, 3, 2, 1, 6]，第一遍循环之后，堆栈中从低到顶剩余的元素是9,8,7,6，此时再从头开始扫描一遍数组，9，发现栈顶是6，所以6的右边更大数是9，弹出6；栈顶是7，7的右边更大数也是9；同理8，弹栈；最后栈顶只剩9了。
代码如下：

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

本代码提交AC，用时119MS。