面试题38：数字在排序数组中出现的次数
题目：统计一个数字在排序数组中出现的次数。例如输入排序数组{1，2，3，3，3，3，4，5}和数字3，由于3在这个数组中出现了4次，因此输出4。
主要的思路是进行二分搜索找到第一个位置和最后一个位置，一开始的代码如下
1 #include <cstdio> 2 #include <cstdlib> 3 4 int num[102]; 5 int n; 6 7 int getFirstk(int from, int to, int k) { 8 if(to < from) { 9 return -1; 10 } 11 12 int mid = (from+to)/2; 13 if(num[mid] < k) { 14 return getFirstk(mid+1,to,k); 15 } 16 else if(num[mid] > k) { 17 return getFirstk(from,mid-1,k); 18 } 19 else { 20 if(mid == 0) { 21 return 0; 22 } 23 if(num[mid-1] == k) { 24 return getFirstk(from,mid-1,k); 25 } 26 else if(num[mid-1] != k) { 27 return mid; 28 } 29 } 30 } 31 32 int getLastK(int from, int to, int k) { 33 if(to < from) { 34 return -1; 35 } 36 37 int mid = (from+to)/2; 38 if(num[mid] < k) { 39 return getLastK(mid+1,to,k); 40 } 41 else if(num[mid] > k) { 42 return getLastK(from, mid-1, k); 43 } 44 else { 45 if(mid == n-1) { 46 return mid; 47 } 48 else { 49 if(num[mid+1] == k) { 50 return getLastK(mid+1,to,k); 51 } 52 else { 53 return mid; 54 } 55 } 56 } 57 } 58 59 int main(int argc, char const *argv[]) 60 { 61 62 while(scanf("%d",&n) != EOF) { 63 for(int i = 0; i < n; i++) { 64 scanf("%d",&num[i]); 65 } 66 int k; 67 scanf("%d",&k); 68 int from = getFirstk(0,n-1,k); 69 int to = getLastK(0,n-1,k); 70 int ans; 71 72 ans = to - from+1; 73 74 printf("%d ",ans); 75 } 76 return 0; 77 }
这段代码的主要问题在于没有考虑元素没有找到的情况，倘若都没找到，from和to都为-1，则会导致结果为1
修改如下
1 #include <cstdio> 2 #include <cstdlib> 3 4 int num[102]; 5 int n; 6 7 int getFirstk(int from, int to, int k) { 8 if(to < from) { 9 return -1; 10 } 11 12 int mid = (from+to)/2; 13 if(num[mid] < k) { 14 return getFirstk(mid+1,to,k); 15 } 16 else if(num[mid] > k) { 17 return getFirstk(from,mid-1,k); 18 } 19 else { 20 if(mid == 0) { 21 return 0; 22 } 23 if(num[mid-1] == k) { 24 return getFirstk(from,mid-1,k); 25 } 26 else if(num[mid-1] != k) { 27 return mid; 28 } 29 } 30 } 31 32 int getLastK(int from, int to, int k) { 33 if(to < from) { 34 return -1; 35 } 36 37 int mid = (from+to)/2; 38 if(num[mid] < k) { 39 return getLastK(mid+1,to,k); 40 } 41 else if(num[mid] > k) { 42 return getLastK(from, mid-1, k); 43 } 44 else { 45 if(mid == n-1) { 46 return mid; 47 } 48 else { 49 if(num[mid+1] == k) { 50 return getLastK(mid+1,to,k); 51 } 52 else { 53 return mid; 54 } 55 } 56 } 57 } 58 59 int main(int argc, char const *argv[]) 60 { 61 62 while(scanf("%d",&n) != EOF) { 63 for(int i = 0; i < n; i++) { 64 scanf("%d",&num[i]); 65 } 66 int k; 67 scanf("%d",&k); 68 int from = getFirstk(0,n-1,k); 69 int to = getLastK(0,n-1,k); 70 int ans; 71 if(from == -1 || to == -1) { 72 ans = 0; 73 } 74 else { 75 ans = to - from+1; 76 } 77 printf("%d ",ans); 78 } 79 return 0; 80 }
找到一个在线测试的oj
http://www.nowcoder.com/books/coding-interviews/70610bf967994b22bb1c26f9ae901fa2?rp=2
此oj的编译条件很严格，而且形式与平常的oj略有不同，花了一点时间来适应，ac代码如下
1 class Solution { 2 public: 3 int GetNumberOfK(vector<int> data ,int k) { 4 int n = data.size(); 5 if(n < 0) { 6 return 0; 7 } 8 int from = getFirstk(data,0,n-1,k); 9 int to = getLastK(data,0,n-1,k,n); 10 if(from == -1 || to == -1) { 11 return 0; 12 } 13 int ans = to - from+1; 14 return ans; 15 } 16 17 int getFirstk(const vector<int> &num,int from, int to, int k) { 18 if(to < from) { 19 return -1; 20 } 21 22 int mid = (from+to)/2; 23 if(num[mid] < k) { 24 return getFirstk(num,mid+1,to,k); 25 } 26 else if(num[mid] > k) { 27 return getFirstk(num,from,mid-1,k); 28 } 29 else { 30 if(mid == 0) { 31 return 0; 32 } 33 if(num[mid-1] == k) { 34 return getFirstk(num,from,mid-1,k); 35 } 36 else if(num[mid-1] != k) { 37 return mid; 38 } 39 } 40 return -1; 41 } 42 43 int getLastK(const vector<int> &num,int from, int to, int k, int n) { 44 if(to < from) { 45 return -1; 46 } 47 48 int mid = (from+to)/2; 49 if(num[mid] < k) { 50 return getLastK(num,mid+1,to,k,n); 51 } 52 else if(num[mid] > k) { 53 return getLastK(num,from, mid-1, k,n); 54 } 55 else { 56 if(mid == n-1) { 57 return mid; 58 } 59 else { 60 if(num[mid+1] == k) { 61 return getLastK(num,mid+1,to,k,n); 62 } 63 else { 64 return mid; 65 } 66 } 67 } 68 return -1; 69 } 70 };