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]
Solutions
My solution
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int size = nums1.size();
for(int i = 1; i <= m; ++i){
nums1[size - i] = nums1[m - i];
}
int k = 0, i = size - m, j = 0;
for(int k = 0; k < m + n; ++k){
int a = i < size ? nums1[i] : INT_MAX;
int b = j < n ? nums2[j] : INT_MAX;
if(a < b){
nums1[k] = a;
++i;
}else{
nums1[k] = b;
++j;
}
}
}
};
Optimal solution
Credit: https://leetcode.com/problems/merge-sorted-array/discuss/29515/4ms-C++-solution-with-single-loop
Idea: process from large to small numbers.
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = m - 1, j = n - 1, k = m + n - 1;
while(j >= 0){
nums1[k--] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
}
}
};
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