[LeetCode]Next Greater Element II

题目描述:

LeetCode 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.

题目大意:

给定一个循环数组(末尾元素的下一个元素为起始元素),输出每一个元素的下一个更大的数字(Next Greater Number)。Next Greater Number是指位于某元素右侧,大于该元素,且距离最近的元素。如果不存在这样的元素,则输出-1。

注意:给定数组长度不超过10000。

解题思路:

解法I 栈(Stack)

时间复杂度O(n)

参考LeetCode Discuss:

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

解法参照LeetCode 496. Next Greater Element I

对于循环数组的处理,将nums数组遍历两次,下标对len(nums)取模

Python代码:

class Solution(object):
    def nextGreaterElements(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        stack = []
        size = len(nums)
        ans = [-1] * size
        for x in range(size * 2):
            i = x % size
            while stack and nums[stack[-1]] < nums[i]:
                ans[stack.pop()] = nums[i]
            stack.append(i)
        return ans

解法II 朴素解法

时间复杂度O(n ^ 2)

Java代码:

public class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int size = nums.length;
        int[] ans = new int[size];
        Arrays.fill(ans, -1);
        for (int i = 0; i < size; i++) {
            for (int j = i + 1; j % size != i; j++) {
                if (nums[j % size] > nums[i]) {
                    ans[i] = nums[j % size];
                    break;
                }
            }
        }
        return ans;
    }
}

 

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