难度Easy
原题连接
内容描述
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]
思路 - 时间复杂度: O(n)- 空间复杂度: O(1)******
由于数组是已经排序好的,我们可以用两个指针 i,j 指向两个数组的第一个元素,当nums[i] < nums[j],++i。否则就 ++j,并把j所指向的元素插入nums1。不过要注意这两个数组的长度。
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = 0,j = 0;
while((i < m) && (j < n))
{
if(nums1[i] < nums2[j])
i++;
else
{
nums1.insert(nums1.begin() + i,nums2[j++]);
i++;
m++;
}
}
if(j < n)
{
nums1.erase(nums1.begin() + m,nums1.end());
nums1.insert(nums1.end(),nums2.begin() + j,nums2.begin() + n);
}
else
nums1.erase(nums1.begin() + m,nums1.end());
}
};