题目不难，关键是边界条件要想清楚。先写一个时间复杂度为O(K) 的解法。
#include <iostream>
using namespace std;
//a[] increase
//b[] decrease
//use ret_value to return the result
//function ret reprsent the error if not 0
int find_k(int a[], int b[], int m, int n, int k, int& ret_value) {
if(k<=0 || m<0 || n<0 || k>(m+n)) {
return -1;
}
int i = 0, x = m-1, y = 0;
while (i<k && x>=0 && y<=n-1) {
if (a[x] > b[y]) {
ret_value = a[x];
i++;
x--;
} else if (a[x] < b[y]) {
ret_value = b[y];
i++;
y++;
} else { //equal
ret_value = a[x];
i+=2;
x--;
y++;
}
}
cout << "x:" << x << endl
<< "y:" << y << endl
<< "i:" << i << endl
<< "v:" << ret_value << endl;
i = k-i;
if (i > 0) {
if (x<0) {
ret_value = a[y+i-1];
} else if (y > n-1) {
ret_value = a[x-i+1];
}
}
return 0;
}
int main () {
int a[] = {1,2,3};
int b[] = {6,5,3};
int k=0;
cout << "please input the k:";
cin >> k;
int value = -1;
find_k(a,b,3,3,k,value);
cout << value << endl;
return 0;
}再附上一个时间复杂服为O(log n)的算法实现，
算法思想是：
每次从两个数组里数 k/2 个数，比较其大小，较大的那个数所在的数组里这 k/2 个数， 肯定在前k大个数里面；
这样就排除了k/2个数，然后再在剩下的数里面找第k/2大的数，循环到找到最后一个数为止。
为了简化一下，假设两个数组都为降序。
注意程序的31~34行，如果不判断相等，当最后还剩一个数的时候，第20行
mid_b = pb + kb -1;
mid_b 会被赋值为一个错误的位置，导致程序陷入死循环
//both a[] b[] decrease
int find_k_both_increase_O_log_k(int a[], int b[], int m, int n, int k, int& ret_value) {
if(k<=0 || m<0 || n<0 || k>(m+n)) {
return -1;
}
int ka = 0;
int kb = 0;
int pa = 0;
int pb = 0;
int mid_a = 0;
int mid_b = 0;
int i = k;
while (i>0 && pa<m && pb<n) {
ka = i/2;
kb = i-ka;
mid_a = pa + ka -1;
mid_b = pb + kb -1;
if (mid_a<m && mid_b<n) {
if (a[mid_a] > b[mid_b]) {
ret_value = a[mid_a];
pa = mid_a + 1;
i = i - ka;
} else if (a[mid_a] < b[mid_b]) {
ret_value = b[mid_b];
pb = mid_b + 1;
i = i - kb;
} else { //equal
ret_value = a[mid_a];
return 0;
}
}
cout << "pa:" << pa << endl
<< "pb:" << pb << endl
<< "i:" << i << endl;
cout <<endl;
}
if (i > 0) {
if (pa > m-1) {
ret_value = a[pb-1+i];
} else if (pb > n-1) {
ret_value = a[pa-1+i];
}
}
return 0;
}在leetcode上碰到该题的改进版本，被标识为hard，看起来简单的题目，其实不简单。
附上题目和代码如下：
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)).
找第k大个数，和我上面一个稍微有点不同，写起来很多边界条件，回头多看几遍。
int find(int A[], int m, int B[], int n, int k) {
if (m==0) {
return B[k-1];
}
if (n==0) {
return A[k-1];
}
int pa=0, pb=0;
int sa=0, sb=0;
int ka=0, kb=0;
while (k>1 && sa<m && sb<n) {
ka = k/2;
kb = k-ka;
pa = sa + ka -1;
pb = sb + kb -1;
if (pa >= m) {
pa = m-1;
ka = m-sa;
kb = k-ka;
pb = sb + kb -1;
} else if (pb >= n) {
pb = n-1;
kb = n-sb;
ka = k - kb;
pa = sa + ka -1;
}
if (A[pa] > B[pb]) {
k = ka;
sb = pb+1;
} else if (A[pa] < B[pb]) {
k = kb;
sa = pa+1;
} else { //A[pa] == B[pb]
return A[pa];
}
}
if (sa>=m) {
return B[sb+k-1];
} else if (sb >= n) {
return A[sa+k-1];
}
if (k == 1) {
return A[sa] < B[sb] ? A[sa] : B[sb];
}
}
class Solution {
public:
double findMedianSortedArrays(int A[], int m, int B[], int n) {
int k=(m+n+1)/2;
if ((m+n)&1) {
return find(A, m, B, n, k);
} else {
return (double)(find(A, m, B, n, k) + find(A, m, B, n, k+1))/2;
}
}
};
