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test03.java
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58 lines (55 loc) · 1.85 KB
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package byteDance;
/**
* Created by lizeyang on 2020/5/12.
* function:合并两个有序数组
*/
public class test03 {
//解法1：归并数组，创建一个新数组然后merge，时间复杂度O(m+n)，空间复杂度O(m+n)
public static int[] test(int[] nums1, int[] nums2) {
if (nums1 == null) {
return nums2;
} else if (nums2 == null) {
return nums1;
} else if (nums1 == null || nums2 == null) {
return null;
}
int[] res = new int[nums1.length + nums2.length];
int i = 0, j = 0, k = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
res[k++] = nums1[i++];
} else if (nums1[i] > nums2[j]) {
res[k++] = nums2[j++];
} else {
res[k++] = nums1[i++];
res[k++] = nums2[j++];
}
}
while (i < nums1.length) {
res[k++] = nums1[i++];
}
while (j < nums2.length) {
res[k++] = nums2[j++];
}
return res;
}
//解法2:利用数组有序特点，从后开始比较，三指针，时间复杂度O(m+n)
public void merge(int[] nums1, int m, int[] nums2, int n) {
int len1 = m - 1;
int len2 = n - 1;
int len = m + n - 1;
while (len1 >= 0 && len2 >= 0) {
nums1[len--] = nums1[len1] > nums2[len2] ? nums1[len1--] : nums2[len2--];
}
//当len1结束时，将nums2数组元素拷贝到nums1中
System.arraycopy(nums2, 0, nums1, 0, len2 + 1);
}
public static void main(String[] args) {
int[] nums1 = {1, 2, 3, 4};
int[] nums2 = {2, 5, 7, 9};
int[] res = test(nums1, nums2);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
}