1. Title Description
Given two positive-ordered (from small to large) arrays of size m and n, nums1 and nums2. Please find and return the median of the two positive ordinal groups.
Example 1:
Input: nums1 = [1,3], nums2 = [2] output: 2.00000 explanation: merge array = [1,2,3], median 2 example 2:
Input: nums1 = [1,2], nums2 = [3,4] output: 2.50000 description: merged array = [1,2,3,4], median (2 + 3) / 2 = 2.5 example 3:
Input: nums1 = [0,0], nums2 = [0,0] output: 0.00000 example 4:
Nums1 = [], nums2 = [1] Output: 1.00000 Example 5:
Input: nums1 = [2], nums2 = [] Output: 2.00000
Source: LeetCode link: leetcode-cn.com/problems/me… Copyright belongs to the Collar buckle network. Commercial reprint please contact official authorization, non-commercial reprint please indicate the source.
Second, complete code
class Solution {
public static void main(String[] args) {
int nums1[] = {0.0.0.0.0};
int nums2[] = {-1.0.0.0.0.0.1};
double median_num = findMedianSortedArrays(nums1, nums2);
System.out.println("Median Number is:" + median_num);
}
public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
double sum = 0.00000;
int num_len1 = nums1.length;
int num_len2 = nums2.length;
if (num_len1 > 1000 || num_len2 > 1000) {
return sum;
}
int merge_num[] = new int[num_len1 + num_len2];
int num1_index = 0;
int num2_index = 0;
int merge_index = 0;
while (num1_index < num_len1 && num2_index < num_len2) {
if (nums1[num1_index] < nums2[num2_index]) {
merge_num[merge_index++] = nums1[num1_index++];
} else if (nums1[num1_index] > nums2[num2_index]) {
merge_num[merge_index++] = nums2[num2_index++];
} else{ merge_num[merge_index++] = nums1[num1_index++]; }}if (num1_index == num_len1) {
while (num2_index < num_len2)
merge_num[merge_index++] = nums2[num2_index++];
} else {
while (num1_index < num_len1)
merge_num[merge_index++] = nums1[num1_index++];
}
if (merge_num.length % 2= =0) {
sum = (merge_num[merge_num.length / 2 - 1] + merge_num[merge_num.length / 2) /2.00000;
} else {
sum = merge_num[(merge_num.length - 1) / 2];
}
returnsum; }}Copy the code