题目描述:给定排序后的字符串数组,中间有一些空串,要求找到给定字符串的位置
思路:
(1)遍历,最慢的
(2)二分查找,当mid处为空串,就找到最近的非空串继续寻找。如果需要找空串?(单独处理)
1 #include <iostream> 2 #include <queue> 3 #include <climits> 4 #include <algorithm> 5 #include <memory.h> 6 #include <stdio.h> 7 #include <ostream> 8 #include <vector> 9 #include <list> 10 #include <cmath> 11 #include <string> 12 #include <stdexcept> 13 #include <stack> 14 #include <map> 15 using namespace std; 16 17 int fun(vector<string> a,string target) 18 { 19 int l = 0; 20 int r = a.size() - 1; 21 while(l <= r) 22 { 23 int mid = (l+r)/2; 24 if(a[mid] == target) 25 { 26 return mid; 27 } 28 else if(a[mid] == "") 29 { 30 int k = mid + 1; 31 while(k <= r) 32 { 33 if(a[k] != "") 34 break; 35 } 36 if(k <= r) 37 { 38 if(a[k] == target) 39 { 40 return k; 41 } 42 else if(a[k] > target) 43 { 44 r = k - 1; 45 } 46 else if(a[k] < target) 47 { 48 l = k + 1; 49 } 50 } 51 else 52 { 53 k = mid - 1; 54 while(k >= l) 55 { 56 if(a[k] != "") 57 break; 58 } 59 if(k >= l) 60 { 61 if(a[k] == target) 62 { 63 return k; 64 } 65 else if(a[k] > target) 66 { 67 r = k - 1; 68 } 69 else if(a[k] < target) 70 { 71 l = k + 1; 72 } 73 } 74 else 75 return -1; 76 } 77 } 78 else if(a[mid] > target) 79 { 80 r = mid - 1; 81 } 82 else if(a[mid] < target) 83 { 84 l = mid + 1; 85 } 86 } 87 } 88 89 int main() 90 { 91 vector<string> input; 92 input.push_back("at"); 93 input.push_back(""); 94 input.push_back(""); 95 input.push_back(""); 96 input.push_back("ball"); 97 input.push_back(""); 98 input.push_back(""); 99 input.push_back("car"); 100 input.push_back(""); 101 input.push_back(""); 102 input.push_back("dad"); 103 input.push_back(""); 104 input.push_back(""); 105 cout<<fun(input,"ball")<<endl; 106 return 0; 107 }