88. Merge Sorted Array [easy] (Python)

题目链接

https://leetcode.com/problems/merge-sorted-array/

题目原文

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

题目翻译

给定两个有序整数数组nums1和nums2，将nums2合并到nums1中得到一个有序数组。
注意：假定nums1有足够的空间（其大小大于等于m+n）来存放nums2的元素。nums1和nums2初始时分别有m和n个元素。

思路方法

思路一

既然要合并到nums1中，则从合并后nums1的尾部元素开始，依次向前确定每个元素的值应该是多少。程序需要用到三个指针。

代码

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        p, q, k = m-1, n-1, m+n-1
        while p >= 0 and q >= 0:
            if nums1[p] > nums2[q]:
                nums1[k] = nums1[p]
                p, k = p-1, k-1
            else:
                nums1[k] = nums2[q]
                q, k = q-1, k-1
        nums1[:q+1] = nums2[:q+1]

思路二

用一个小小的trick，只用两个指针即可。

代码

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        p, q = m-1, n-1
        while p >= 0 and q >= 0:
            if nums1[p] > nums2[q]:
                nums1[p+q+1] = nums1[p]
                p = p-1
            else:
                nums1[p+q+1] = nums2[q]
                q = q-1
        nums1[:q+1] = nums2[:q+1]

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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