题目
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
简单来说就是求两个已经排好序的数组的中位数
代码
1. 先合并两个数组进行排序，然后求中位数
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
//先给两个数组进行排序
int[] datas = new int[nums1.length + nums2.length];
int i = 0;
int k = 0;
int index = 0;
while (i<nums1.length || k<nums2.length) {
if(k>=nums2.length) {
datas[index++] = nums1[i++];
continue;
}
if(i>=nums1.length) {
datas[index++] = nums2[k++];
continue;
}
if(nums1[i]<nums2[k]) {
datas[index++] = nums1[i++];
}else {
datas[index++] = nums2[k++];
}
}
//取中位数
if(datas.length%2==0) {
//取中间两个数，求平均数
return (datas[datas.length/2 -1] + datas[datas.length/2])/2.0;
}else {
//取中间一个数
if(datas.length == 1) {
return datas[0];
}
return datas[datas.length/2];
}
}
2. 已经排好了序，直接按取第几个数这样
public double findnum2(int[] nums1,int[] nums2) {
int m = nums1.length, n = nums2.length;
int i = 0, j = 0, index = 0, len = m + n, loop = len / 2;
int mid0 = 0, mid1 = 0;
while (index <= loop) {
mid0 = mid1;
if (i < m && j < n) {
mid1 = nums1[i] < nums2[j] ? nums1[i++] : nums2[j++];
} else if (i < m) {
mid1 = nums1[i++];
} else {
mid1 = nums2[j++];
}
++index;
}
if (len % 2 == 0) {
return (mid0 + mid1) / 2.0;
} else {
return mid1;
}
}
