Median

http://pat.zju.edu.cn/contests/pat-a-practise/1029

二分搜索median。

代码一： 时间复杂度log(m  + n)。

思路：比较两个数组的中间元素，然后可以排除掉其中一个数组的一半。

假设数组A, B, 我们要找的是A, B合并为一个数组后的第median个元素。令A, B中间元素分别为A[mid1], B[mid2]：

如果 A[mid1] == B[mid2]，则结果就为A[mid1];

如果 A[mid1] < B[mid2]，这时B[mid2]的前面一共有 mid1 + mid2 + 1个数字，

此时需要判断mid1 + mid2 + 1和median 的大小：

如果前者大，意味着我们要找的数字就在这 mid1 + mid2 + 1个数字中，从而我们可以排除B数组中的后一半；

如果后者大，意味着我们要找的数字不在这 mid1 + mid2 + 1个数字中，从而我们可以排除A 数组中的前一半。

当A[mid1] > B[mid2]时情况类似。

#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <algorithm>#include <stack>using namespace std;long long *a, *b;long long findMedian(long long *v1, long long *v2, int len1, int len2, int indexOfMid){if(len1 == 0) return v2[indexOfMid - 1];if(len2 == 0) return v1[indexOfMid - 1];int mid1 = (len1 - 1) >> 1, mid2 = (len2 - 1) >> 1;int index = mid1 + mid2 + 1;if(v1[mid1] == v2[mid2]) return v1[mid1];if(v1[mid1] < v2[mid2]){if(index < indexOfMid) return findMedian(v1 + mid1 + 1, v2, len1 - mid1 - 1, len2, indexOfMid - mid1 - 1);else return findMedian(v1, v2, len1, mid2, indexOfMid);}else{if(index < indexOfMid) return findMedian(v1, v2 + mid2 + 1, len1, len2 - mid2 - 1, indexOfMid - mid2 - 1);return findMedian(v1, v2, mid1, len2, indexOfMid);}}int main(){int m, n, i;long long x;scanf("%d", &m);a = new long long[m + 10];for(i = 0; i < m; i ++){scanf("%lld", &a[i]);}scanf("%d", &n);b = new long long[n + 10];for(i = 0; i < n; i ++){scanf("%lld", &b[i]);}int mid = (m + n + 1) >> 1;long long ans = findMedian(a, b, m, n, mid);printf("%lld\n", ans);delete a;delete b;return 0;}

代码二：时间复杂度logm*logn(想复杂了)

#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <algorithm>#include <stack>using namespace std;long long *a, *b;int binaryS1(long long *v, long long target, int size) // find first smaller than target{int start = 0, end = size - 1;while(start < end){int mid = (start + end) >> 1;if(v[mid] >= target) end = mid - 1;else start = mid + 1;}if(v[start] >= target) start -- ;return start;}int binaryS2(long long *v, long long target, int size) // find first larger than target{int start = 0, end = size - 1;while(start < end){int mid = (start + end) >> 1;if(v[mid] <= target) start = mid + 1;else end = mid - 1;}if(v[start] <= target) start ++ ;return start;}bool isFind = false;long long findMedian(long long *v1, long long *v2, int size1, int size2){int start = 0, end = size1 - 1;int median = (size1 + size2) >> 1;if((size1 + size2) % 2 == 1) median ++ ;while(start <= end){int mid = (start + end) >> 1;int index1 = binaryS1(v2, v1[mid], size2);int index2 = binaryS2(v2, v1[mid], size2);int posMin = mid + 1 + index1 + 1;int posMax = mid + 1 + index2;if(posMin <= median && posMax >= median) {isFind = true;return v1[mid];}else if(posMin > median) end = mid - 1;else start = mid + 1;}return -1;}int main(){int m, n, i;long long x;scanf("%d", &m);a = new long long[m + 10];for(i = 0; i < m; i ++){scanf("%lld", &a[i]);}scanf("%d", &n);b = new long long[n + 10];for(i = 0; i < n; i ++){scanf("%lld", &b[i]);}long long ans = findMedian(a, b, m, n);if(isFind == false) ans = findMedian(b, a, n, m);printf("%lld\n", ans);delete a;delete b;return 0;}

当然这题还有O(m + n)的算法，即归并排序思想。