题目描述:(困难难度)
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)).
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
题目解释:
有长度为m的整形数组1和长度为n的整形数组2。找到这两个数组的中间值。总的运行时间复杂度应该是O(log(m+n))。
题目解法:
1.我的解法。我的解法比较笨,先合并两个数组,然后取中间的数,最后返回值(其实没有必要全部合并,只需要查到相应的位置就可以了,后续的操作没有任何必要,当然啦,这两种的时间复杂度都是O(m+n),没有达到要求)。代码如下:
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int length = nums1.length + nums2.length;
int[] temp = new int[length];
int work1 = 0;
int work2 = 0;
for(int i = 0; i < length ; i++) {
if(work1 < nums1.length && work2 < nums2.length) {
if(nums1[work1] < nums2[work2]) {
temp[i] = nums1[work1];
work1 ++;
} else {
temp[i] = nums2[work2];
work2 ++;
}
} else if(work1 >= nums1.length && work2 < nums2.length) {
temp[i] = nums2[work2];
work2 ++;
} else if(work1 < nums1.length && work2 >= nums2.length) {
temp[i] = nums1[work1];
work1 ++;
}
}
if(length % 2 == 1) {
return Double.valueOf(temp[length / 2]);
} else {
return (temp[length / 2] + temp[length / 2 - 1]) / 2.0;
}
}
}