Merge Sorted Array
Easy
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]
Analysis
这一题的trick在于,因为是数组,不适合插入操作,所以从nums1[]的末端开始反向填充,就可以避免数据大量搬运。
Solution
One pass - O(n) time, O(1) space - (2ms, 100%)
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int k = m + n;
int i = m - 1;
int j = n - 1;
for (int l = k - 1; l >= 0; l--) {
// nums2[] elements have all been copied
if (j < 0) {
return;
}
// nums1[] elements have all been copied
if (i < 0) {
nums1[l] = nums2[j];
j--;
continue;
}
// compare and assign elements from nums1, nums2
if (nums2[j] > nums1[i]) {
nums1[l] = nums2[j];
j--;
} else {
nums1[l] = nums1[i];
i--;
}
}
}
}
// [1,2,3,0,0,0]
// [4,5,6]
// [4,5,6,0,0,0]
// [1,2,3]
Another more concise implementation (2ms, 100%)
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int k = m + n - 1;
int i = m - 1;
int j = n - 1;
while (i >= 0 && j>= 0) {
// compare and assign elements from nums1, nums2
if (nums2[j] > nums1[i]) {
nums1[k--] = nums2[j--];
} else {
nums1[k--] = nums1[i--];
}
}
while (j >= 0) {
nums1[k--] = nums2[j--];
}
}
}
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