[LeetCode]496. Next Greater Element I

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
    For number 1 in the first array, the next greater number for it in the second array is 3.
    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

难度

Easy

方法1

直接按照题目意思查找，时间复杂度O(m*n)

python代码

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        result = []
        for findNum in findNums:
            numsCount = len(nums)
            i = 0
            equalFlag = False
            while i < numsCount:
                if equalFlag:
                    if findNum < nums[i]:
                        result.append(nums[i])
                        break;
                    else:
                        i += 1
                else:
                    if findNum == nums[i]:
                        equalFlag = True
                    i += 1
            if i == numsCount:
                result.append(-1)

        return result

assert Solution().nextGreaterElement([4,1,2], [1,3,4,2]) == [-1,3,-1]
assert Solution().nextGreaterElement([2,4], [1,2,3,4]) == [3,-1]

方法2

利用map和stack，遍历nums2，将每个元素和它next greater element的对应关系存入map中，最后遍历nums1从map中取值，如果没有返回-1。时间复杂度O(m+n)

python代码

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        numsStack = []
        numsMap = {}

        for num in nums:
            while len(numsStack) and (numsStack[-1] < num):
                numsMap[numsStack.pop()] = num
            numsStack.append(num)

        result = []
        for findNum in findNums:
            result.append(numsMap.get(findNum, -1))

        return result

assert Solution().nextGreaterElement([4,1,2], [1,3,4,2]) == [-1,3,-1]
assert Solution().nextGreaterElement([2,4], [1,2,3,4]) == [3,-1]