Description
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
Note:
The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.
Example:
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
描述
- 题意是给定了两个排好序的数组，让把这两个数组合并,不要使用额外的空间，把第二个数组放到第一个数组之中.
- 两个数组nums1，长度m，数组nums2，长度n，我们从两个数组的末尾，p，q开始比较，我们把较大的数放在p+q+1的位置.
- 如此循环，最后把nums2剩下的元素放到nums1即可.
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2018-12-25 17:10:16 # @Last Modified by: 何睿 # @Last Modified time: 2018-12-25 17:10:16 class Solution: 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分别指向数组的最后一个位置. p, q = m-1, n-1 # 当数组不为空的才进行循环. while p >= 0 and q >= 0: # 把较大的数放在末尾 if nums1[p] > nums2[q]: nums1[p+q+1] = nums1[p] p = p-1 else: nums1[p+q+1] = nums2[q] q = q-1 # 把nums2剩下的元素放在nums1中. nums1[:q+1] = nums2[:q+1] if __name__ == "__main__": so = Solution() nums1 = [1, 2, 3, 0, 0, 0] nums2 = [2, 5, 6] so.merge(nums1, 3, nums2, 3) print(nums1)
源代码文件在这里.
