LeetCode 88. Merge Sorted Array.jpg
LeetCode 88. Merge Sorted Array
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)
