Search in Rotated Sorted Array II

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Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

这道题目尽管用了 binary search，但 time complexity 依然是 n, 但是还是用binary search 实现了一下：

修改过的代码，比 “search in rotated sorted array” 只多了两个while 循环，但是也是反复修改才添加正确的，面试的时候不一定能一次写正取。

</pre><pre name="code" class="java">public class Solution {    public boolean search(int[] A, int target) {        if (A == null || A.length == 0) {            return false;        }        int start = 0;        int end = A.length - 1;        int mid;        boolean inLeft;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if (target == A[mid]) {                return true;            } else {                /**                 * 当while 循环被执行时，说明至少有一边的数据是完全相等的，例如 1121111，或者211111                 * 对于112111，执行完两个while循环以后，start 和 end 都指向 2                 * 则跳出最外层while循环                 *                  * 对于211111这种，执行以后end ＝＝ mid，继续操作                 * 所以两个while循环主要用来处理这种情况                 * 处理完以后，就可以按照 “search in rotated sorted array ” 的方法来写                 */                while (start< A.length-1 && A[mid] == A[start]) {                    start ++;                }                while (end > 0 && A[mid] == A[end]) {                    end--;                } if (target < A[mid]) {                        if (A[mid] > A[end] && A[end] >= target) {                            start = mid + 1;                        } else {                             end = mid - 1;                        }                } else {                        if (A[mid] < A[start] && target >= A[start]) {                            end = mid - 1;                        } else {                            start = mid + 1;                         }                }                            }        }        if (target == A[start] || target == A[end]) {            return true;        }        return false;    }}

第一次写的答案：

public class Solution {    public boolean search(int[] A, int target) {        if (A == null || A.length == 0) {            return false;        }        int start = 0;        int end = A.length - 1;        int mid;        boolean inLeft;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if (target == A[mid]) {                return true;            } else {                if (A[mid] == A[start] && A[mid] == A[end]) {                    int i;                    for (i = mid - 1; i >= 0; i--) {                        if (A[i] != A[mid]) {                            end = i ;                            break;                        }                     }                     if (end != i) {                         start = mid + 1;                     }                } else if (A[mid] == A[start] && A[mid] != A[end]){                    start = mid + 1;                }else if (A[mid] != A[start] && A[mid] == A[end]){                    end = mid - 1;                } else {                    if (target < A[mid]) {                        if (A[mid] > A[end] && A[end] >= target) {                            start = mid + 1;                        } else {                             end = mid - 1;                        }                    } else {                        if (A[mid] < A[start] && target >= A[start]) {                            end = mid - 1;                        } else {                            start = mid + 1;                         }                    }                }            }        }        if (target == A[start] || target == A[end]) {            return true;        }        return false;    }}

别人的答案（我自己觉得第一份代码比这份答案看上去还要清晰一点）

public class Solution {    public boolean search(int[] A, int target) {        int start = 0, end = A.length-1;        while(end - start >= 2) {        int middle = start + (end-start)/2;        if(A[middle] == target) return true;        if(A[start] < A[middle]) {        if(A[start] <= target && target < A[middle]) end = middle-1;        else start = middle+1;        }        if(A[middle] < A[end]) {        if(A[middle] < target && target <= A[end]) start = middle+1;        else end = middle-1;        }        if(A[start] == A[middle]) start++;        if(A[middle] == A[end]) end--;                }        for(int i = start; i <= end; i++) {        if(A[i] == target) return true;        }        return false;    }}

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