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  • [LeetCode] 503. Next Greater Element II 下一个较大的元素 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.

    496. Next Greater Element I 的拓展,这里的数组是循环的,某一个元素的下一个较大值可以在其前面。

    解法:栈,与496的不同是:循环是2倍的数组长度,用栈来保存降序序列的index。

    Java:

    public int[] nextGreaterElements(int[] nums) {
            int n = nums.length, next[] = new int[n];
            Arrays.fill(next, -1);
            Stack<Integer> stack = new Stack<>(); // index stack
            for (int i = 0; i < n * 2; i++) {
                int num = nums[i % n]; 
                while (!stack.isEmpty() && nums[stack.peek()] < num)
                    next[stack.pop()] = num;
                if (i < n) stack.push(i);
            }   
            return next;
        }
    

    Python:

    def nextGreaterElements(self, nums):
            stack, res = [], [-1] * len(nums)
            for i in range(len(nums)) * 2:
                while stack and (nums[stack[-1]] < nums[i]):
                    res[stack.pop()] = nums[i]
                stack.append(i)
            return res 

    C++:

    vector<int> nextGreaterElements(vector<int>& nums) {
            int n = nums.size();
            vector<int> next(n, -1);
            stack<int> s; // index stack
            for (int i = 0; i < n * 2; i++) {
                int num = nums[i % n]; 
                while (!s.empty() && nums[s.top()] < num) {
                    next[s.top()] = num;
                    s.pop();
                }
                if (i < n) s.push(i);
            }   
            return next;
        } 

      

    类似题目:

    [LeetCode] 496. Next Greater Element I 下一个较大的元素 I

    [LeetCode] 556. Next Greater Element III 下一个较大的元素 III

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/9847937.html
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