LeetCode——Median of Two Sorted Arrays

There are two sorted arrays A and B 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)).

有两个已排序的数组A和B，大小为m 和 n。

找出两数组的中位数

时间复杂度应该为 O(log (m+n))。

简单粗暴的方法就是把两个数组合并成一个数组，排序，取中位数。但是用不到所给的已排序，也体现不了任何算法的意义。时间复杂度：O((m+n)log(m+n))。(不过此方法可以通过。)

public static double findMedianSortedArrays(int A[], int B[]) {int AB[] = new int[A.length + B.length];for (int i = 0; i < A.length; i++) {AB[i] = A[i];}for (int i = 0; i < B.length; i++) {AB[A.length + i] = B[i];}Arrays.sort(AB);double median = AB.length % 2 == 1 ? AB[AB.length >> 1] : (AB[(AB.length - 1) >> 1] + AB[AB.length >> 1]) / 2.0;return median;}

另一种思路：将此问题转换为在有序数组中寻找第K小的数的问题。直接抄了这里的实现。

public static double findMedianSortedArrays(int[] a, int[] b) {int m = a.length, n = b.length;int median = (m + n) / 2;if ((m + n) % 2 == 1) // Single medianreturn findKthElement(a, b, median);else// Average of two mediansreturn (findKthElement(a, b, median) + findKthElement(a, b,median - 1)) / 2.0;}private static double findKthElement(int[] a, int[] b, int k) {int m = a.length, n = b.length;int aStart = 0, bStart = 0; // Keep track of the starting index of both// arrays under considerationint ka, kb;while (true) {if (aStart == m) // a is empty; find the k-th element in breturn b[bStart + k];if (bStart == n) // b is empty; find the k-th element in areturn a[aStart + k];if (k == 0) // Find the smallest elementreturn Math.min(a[aStart], b[bStart]);// Set the indices of the elements under considerationif (k == 1) { // A special case when k/2-1 < 0ka = aStart;kb = bStart;} else { // Make sure the indices are within the boundarieska = Math.min(m - 1, aStart + k / 2 - 1);kb = Math.min(n - 1, bStart + k / 2 - 1);}// Adjust k and the start index according to the relationship// between a[ka] and b[kb]if (a[ka] <= b[kb]) { // Any Element in a[aStart...ka] cannot be the onek -= ka - aStart + 1;aStart = ka + 1;} else { // Any Element in b[bStart...kb] cannot be the onek -= kb - bStart + 1;bStart = kb + 1;}}}