题目描述:
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
代码解决:
package com.jack.algorithm;
/**
* create by jack 2018/10/20
*
* @auther jack
* @date: 2018/10/20 17:38
* @Description:
* 求两个有序数组的中位数
*/
public class MedianOfTwoSortedArrays {
/**
* 题目描述:
* https://leetcode.com/problems/median-of-two-sorted-arrays/
* @param nums1
* @param nums2
* @return
*/
public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
int length1 = nums1.length;
int length2 = nums2.length;
int length = length1+length2;
int [] a = new int [length];
double rs = 0;
if (length1 == 0 && length2 >0) {
for (int i=0;i<length2;i++) {
a[i] = nums2[i];
}
} else if (length1 > 0 && length2 == 0) {
for (int i=0;i<length1;i++) {
a[i] = nums1[i];
}
} else if (length1 > 0 && length2 > 0) {
int i = 0;
int j = 0;
for (int k=0;k<length;k++) {
if (i >= length1 && j < length2) {
a[k] = nums2[j];
j++;
continue;
}
if (i < length1 && j >= length2) {
a[k] = nums1[i];
i++;
continue;
}
if (nums1[i] <= nums2[j]) {
if (i < length1) {
a[k] = nums1[i];
i++;
}
} else {
if (j < length2) {
a[k] = nums2[j];
j++;
}
}
}
}
if (length % 2 == 0) {
rs = (a[length/2-1]+a[length/2])/2.0;
} else {
rs = a[length/2];
}
return rs;
}
public static void main(String[] args) {
int [] nums1 = new int[]{1,2};
//int [] nums2 = new int[]{3,4};
int [] nums2 = new int[]{3,4};
double d = findMedianSortedArrays(nums1,nums2);
System.out.println("中位数为:"+d);
}
}
github地址: