You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of 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: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
Example 2
Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].
Example 3
Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
Constraints
- nums1.length == m + n
- nums2.length == n
- 0 <= m, n <= 200
- 1 <= m + n <= 200
- -10e9 <= nums1[i], nums2[j] <= 10e9
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
Solution
查找插入，然后加一个标识符，前面插不进就直接尾附加。
void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){
int i, j, k, flag;
for (i = 0; i < n; i++) {
flag = 0;
for (j = 0; j < m + i; j++) {
if (nums2[i] < nums1[j]) {
for (k = m + i; k > j; k--) {
nums1[k] = nums1[k - 1];
}
nums1[j] = nums2[i];
flag = 1;
break;
}
}
if (flag == 0) nums1[m + i] = nums2[i];
}
}
从后面操作会更好，下次尝试。
