1. Description of the topic
You are given two integer arrays nums1 and nums2 in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2, respectively.
Please merge nums2 into nums1 so that the merged arrays are also arranged in non-decreasing order.
Note: Ultimately, the merged array should not be returned by the function, but stored in the array nums1. To cope with this situation, the initial length of nums1 is m + n, where the first m elements represent the elements that should be merged, and the last n elements are 0 and should be ignored. The length of nums2 is n.
Example 1:
Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5 ,6]
Explanation: Need to combine [1,2,3] and [2,5,6].
The combined result is [1,2,2,3,5,6], where the elements in nums1 are marked in bold italics.
Example 2:
Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: Need to combine [1] and [].
The combined result is [1].
Example 3:
Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays to be merged are [] and [1].
The combined result is [1].
Note that since m = 0, there are no elements in nums1. The only 0 in nums1 is to ensure that the merged result can be stored in nums1 smoothly.
2. Ideas
The double pointer method
uses the pointers n and m to represent the subscripts of nums2 of nums1, respectively. The sizes of the two array elements are compared from the end of the array, and the corresponding values are assigned from the end of nums1 after the comparison.
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
//双指针法
int index = m+n-1;
while(n >= 1 && m >= 1){
//当前比较 nums2 > nums1
if(nums2[n - 1] >= nums1[ m -1 ]){
nums1[index] = nums2[n - 1];
n--;
}
//当前比较 nums1 > nums2
else{
nums1[index] = nums1[m - 1];
m--;
}
index--;
}
//当 nums2 中还有元素时
if(m == 0){
while(n >= 1){
nums1[index--] = nums2[n - 1];
n--;
}
}
}
};
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