《Cracking the Coding Interview》——第11章：排序和搜索——题目3

2014-03-21 20:55

题目：给定一个旋转过的升序排序好的数组，不知道旋转了几位。找出其中是否存在某一个值。

解法1：如果数组的元素都不重复，那么我的解法是先找出旋转的偏移量，然后进行带偏移量的二分搜索。两个过程都是对数级的。

代码：

 // 11.3 Given a sorted array rotated by a few positions, find out if a value exists in the array.
 // Suppose all elements in the array are unique.
 #include <algorithm>
 #include <vector>
 #include <cstdio>
 using namespace std;

 int rotatedBinarySearch(vector<int> &v, int n, int key)
 {
     int offset;

     if ((int)v.size() < n || n <= ) {
         return -;
     }

     int ll, rr, mm;

     if (v[] < v[n - ]) {
         offset = ;
     } else {
         ll = ;
         rr = n - ;
         while (rr - ll > ) {
             mm = (ll + rr) / ;
             if (v[mm] > v[ll]) {
                 ll = mm;
             } else {
                 rr = mm;
             }
         }
         offset = rr;
     }

     ll = ;
     rr = n - ;
     while (ll <= rr) {
         mm = (ll + rr) / ;
         if (key < v[(mm + offset) % n]) {
             rr = mm - ;
         } else if (key > v[(mm + offset) % n]) {
             ll = mm + ;
         } else {
             return (mm + offset) % n;
         }
     }
     return -;
 }

 int main()
 {
     int n;
     int i;
     vector<int> v;

     while (scanf("%d", &n) ==  && n > ) {
         v.resize(n);
         for (i = ; i < n; ++i) {
             scanf("%d", &v[i]);
         }
         scanf("%d", &i);
         printf("%d\n", rotatedBinarySearch(v, n, i));
     }

     return ;
 }

解法2：如果数组的元素可能存在重复，那么我的思路仍然是先二分查找找出偏移量，然后执行带偏移量的二分搜索。不过，在找偏移量的过程中可能会出现无法决定向左还是向右的情况，比如这两个例子{1, 3, 1, 1, 1}{1, 1, 1, 3, 1}，在第一次二分时，左中右的元素都是‘1’，无法确定应该往哪边走。这时就得扫描整段，{1, 1, 1}全部是同一元素，{1, 3, 1}存在不同元素，所以应该选择{1, 3, 1}进行二分，因为在首尾相同的情况下，中间如果有不同元素的话，表示旋转的偏移量应该会落在这个区间里。找到偏移量以后，之后的查找就是严格二分的了。

代码：

 // 11.3 Given a sorted array rotated by a few positions, find out if a value exists in the array.
 // Suppose the array may contain duplicates, what's it gonna be then?
 #include <algorithm>
 #include <vector>
 #include <cstdio>
 using namespace std;

 int rotatedBinarySearch(vector<int> &v, int n, int key)
 {
     int offset;

     if ((int)v.size() < n || n <= ) {
         return -;
     }

     int ll, rr, mm;
     int i;

     ll = ;
     rr = n - ;
     while (rr - ll >  && v[ll] == v[rr]) {
         mm = (ll + rr) / ;
         if (v[mm] > v[ll]) {
             ll = mm;
             break;
         } else if (v[mm] < v[ll]) {
             rr = mm;
             break;
         } else {
             for (i = ll; i < mm - ; ++i) {
                 if (v[i] != v[i + ]) {
                     break;
                 }
             }
             if (i < mm - ) {
                 rr = mm;
                 break;
             }
             for (i = mm; i < rr - ; ++i) {
                 if (v[i] != v[i + ]) {
                     break;
                 }
             }
             if (i < rr - ) {
                 break;
             }

             // if all elements are the same, it ends here
             return (v[] == key) ?  : -;
         }
     }

     if (v[ll] < v[rr]) {
         offset = ;
     } else {
         // here it is guaranteed v[ll] != v[rr]
         while (rr - ll > ) {
             mm = (ll + rr) / ;
             if (v[mm] >= v[ll]) {
                 ll = mm;
             } else {
                 rr = mm;
             }
         }
         offset = rr;
     }

     // the binary search part remains the same, difference lies in how we find the 'offset'.
     ll = ;
     rr = n - ;
     while (ll <= rr) {
         mm = (ll + rr) / ;
         if (key < v[(mm + offset) % n]) {
             rr = mm - ;
         } else if (key > v[(mm + offset) % n]) {
             ll = mm + ;
         } else {
             return (mm + offset) % n;
         }
     }

     return -;
 }

 int main()
 {
     int n;
     int i;
     vector<int> v;

     while (scanf("%d", &n) ==  && n > ) {
         v.resize(n);
         for (i = ; i < n; ++i) {
             scanf("%d", &v[i]);
         }
         scanf("%d", &i);
         printf("%d\n", rotatedBinarySearch(v, n, i));
     }

     return ;
 }