題目

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)).

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

翻譯

有2個排序過的數(shù)組nums1，nums2，它們的長度分別是m和n。
找到兩個數(shù)組的中位數(shù)，總的運行時間需要是 O(log (m+n))。
例1:

nums1 = [1, 3] nums2 = [2]中位數(shù)：2.0

例2:

nums1 = [1, 2] nums2 = [3, 4]中位數(shù)： (2 + 3)/2 = 2.5

分析

以下解法LeetCode時間為65ms，雖然Accepted了，但時間復(fù)雜度沒到O(log (m+n))，該解法待后續(xù)優(yōu)化。

Java版代碼（時間復(fù)雜度O(m+n),空間復(fù)雜度O(m+n)）:

public class Solution {public double findMedianSortedArrays(int[] nums1, int[] nums2) {if (nums1.length == 0 && nums2.length == 0) {return 0.0;} else if (nums1.length == 0) {return getMid(nums2);} else if (nums2.length == 0) {return getMid(nums1);}Double mid1 = getMid(nums1), mid2 = getMid(nums2);int len = nums1.length + nums2.length;int index1 = len / 2, index2 = len / 2;if (len % 2 == 0) index1--;int sum = 0, current, j = 0, k = 0;if (mid1 < mid2) {j = nums1.length < 2 ? 0 : nums1.length / 2 - 1;} else {k = nums2.length < 2 ? 0 : nums2.length / 2 - 1;}for (int i = j + k; i < len; i++) {if (j >= nums1.length) {if (sum == 0) {sum = nums2[index1 - nums1.length];}return (sum + nums2[index2 - nums1.length]) / 2.0;} else if (k >= nums2.length) {if (sum == 0) {sum = nums1[index1 - nums2.length];}return (sum + nums1[index2 - nums2.length]) / 2.0;}if (nums1[j] < nums2[k]) {current = nums1[j];j++;} else {current = nums2[k];k++;}if (i == index1) {if (index1 == index2) {return current;}sum = current;}if (i == index2) {sum += current;return sum / 2.0;}}return 0;}public Double getMid(int n[]) {int len = n.length;if (len == 0) {return null;}if (len % 2 == 0) {return (n[len / 2 - 1] + n[len / 2]) / 2.0;} else {return n[len / 2] / 1.0;}} }

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