median-of-two-sorted-arrays

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题目描述

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)).

分析：

两种思路：
1. 直接 merge 两个数组，然后求中位数，能过，不过复杂度是 O(n + m)。
2. 用二分的思路去做，这不好想，还要考虑到奇偶。可以转化思维，去求两个有序数组中的第 K 大数，这样就比较好想了。

代码：

1. Merge

java:

package leetcode.easy.easy;public class FindMedianSortedArrays {/** * @param args */public static void main(String[] args) {int[] A = new int[] {};int[] B = new int[] { 2, 3 };System.out.println(findMedianSortedArrays(A, B));}public static double findMedianSortedArrays(int A[], int B[]) {int lenA = A.length;int lenB = B.length;int[] C = new int[lenA + lenB];int i = 0, j = 0, k = 0;if (lenA != 0 && lenB != 0) {while (i < lenA || j < lenB) {if (i == lenA) {C[k++] = B[j++];continue;}if (j == lenB) {C[k++] = A[i++];continue;}if (A[i] > B[j]) {C[k++] = B[j++];} else {C[k++] = A[i++];}}} else {if (lenA == 0) {while (j < lenB) {C[k++] = B[j++];}}if (lenB == 0) {while (i < lenA) {C[k++] = A[i++];}}}if ((lenA + lenB) % 2 == 1) {return C[(lenA + lenB) / 2];}return (C[(lenA + lenB)/2 - 1] + C[(lenA + lenB)/2]) / 2.0;}}

C++:

class Solution {public:double findMedianSortedArrays(int A[], int m, int B[], int n) {vector<int> C;int pa = 0, pb = 0;// point of A & Bwhile (pa < m || pb < n) {if (pa == m) {C.push_back(B[pb++]);continue;}if (pb == n) {C.push_back(A[pa++]);continue;}if (A[pa] > B[pb])C.push_back(B[pb++]);elseC.push_back(A[pa++]);}if ((n + m)&1)return C[(n+m)/2];elsereturn (C[(n+m)/2 - 1] + C[(n+m)/2]) / 2.0;}};

2. 二分

java:

暂时没看懂，待补充

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