【LeetCode】Search in Rotated Sorted Array II 解题报告

【题目】

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.

【解析】

相比Search in Rotated Sorted Array，在递归时需要先去重，再迭代（递归）。

【C++代码】

class Solution {public:    bool search(int A[], int n, int key) {        if (n<=0) return false;        if (n==1){            return (A[0]==key) ? true : false;        }                int low=0, high=n-1;        while( low<=high ){                if (A[low] < A[high] && ( key < A[low] || key > A[high]) ) {                 return false;            }                        //if dupilicates, them binary search the duplication            while (low < high && A[low]==A[high]){                low++;            }                int mid = low + (high-low)/2;            if (A[mid] == key) return true;                //the target in non-rotated array            if (A[low] < A[mid] && key >= A[low] && key< A[mid]){                high = mid - 1;                continue;            }            //the target in non-rotated array            if (A[mid] < A[high] && key > A[mid] && key <= A[high] ){                low = mid + 1;                continue;            }            //the target in rotated array            if (A[low] > A[mid] ){                high = mid - 1;                continue;            }            //the target in rotated array            if (A[mid] > A[high] ){                low = mid + 1;                continue;            }                        //reach here means nothing found.            low++;        }        return false;    }};

【简洁Java版】

public class Solution {    public boolean search(int[] A, int target) {        int l = 0;        int r = A.length - 1;        while (l <= r) {            while (l < r && A[l] == A[r]) l++;            int mid = (l + r) / 2;            if (target == A[mid]) return true;            if (A[l] <= A[r]) {                if (target < A[mid]) r = mid - 1;                else l = mid + 1;            } else if (A[l] <= A[mid]) {                if (target < A[l] || target > A[mid]) l = mid + 1;                else r = mid - 1;            } else {                if (target < A[mid] || target > A[r]) r = mid - 1;                else l = mid + 1;            }        }        return false;    }}