1+ /*题目:
2+ There are two sorted arrays nums1 and nums2 of size m and n respectively.
3+
4+ Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
5+
6+ Example 1:
7+ nums1 = [1, 3]
8+ nums2 = [2]
9+
10+ The median is 2.0
11+ Example 2:
12+ nums1 = [1, 2]
13+ nums2 = [3, 4]
14+
15+ The median is (2 + 3)/2 = 2.5
16+
17+ */
18+ 19+ /*解析:
20+
21+ 要想找到中位数,就必须知道中位数的概念。从小到大排序的一列数,取其中的中间的数,如果长度是偶数的话取中间的两个数/2
22+ 所以最重要,最核心的是:排序
23+
24+ 如何将两个有序的数组,合并成一个有序的数组呢?
25+ 算法:顺序扫描,每次将较小的元素放进结果数组中,和两个链表相加的题有些类似,可以复习一下之前的。这题不难。
26+
27+ 复杂度:O(n)
28+
29+ */
30+ 31+ class Solution {
32+ public double findMedianSortedArrays (int [] nums1 , int [] nums2 ) {
33+ int [] result = new int [nums1 .length +nums2 .length ];
34+ double median ;
35+ //将数组1作为标准
36+ int i = 0 ;//数组1的临时变量
37+ int j = 0 ;//数组2的临时变量
38+ int count = 0 ;//结果数组的临时变量
39+ for ( i = 0 ;i <nums1 .length &&j <nums2 .length ;) {
40+ //将两者最小的数,放入目标数组
41+ if (nums1 [i ]>nums2 [j ]) {
42+ result [count ++]=nums2 [j ++];
43+ }
44+ else {
45+ result [count ++]=nums1 [i ++];
46+ }
47+ }
48+ //数组还有元素没有排好
49+ while (i <nums1 .length ){
50+ result [count ++]=nums1 [i ++];
51+ }
52+ 53+ while (j <nums2 .length ) {
54+ result [count ++]=nums2 [j ++];
55+ }
56+ 57+ //进行计算中位数
58+ if (result .length %2 ==0 ) {
59+ median =(result [result .length /2 ]+result [result .length /2 -1 ])/2.0 ;
60+ 61+ }
62+ else {
63+ median = result [result .length /2 ];
64+ }
65+ 66+ return median ;
67+ }
68+ }
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