"Leetcode" Search in rotated Sorted Array II Problem Solving report

Topic

Follow to "Search in rotated Sorted Array":
What if duplicates is allowed?

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

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

Resolution

Compared to the search in rotated Sorted Array, it is necessary to go first, then iterate (recursively) in recursion.

"C + + code"

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] && ke Y <= 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; }};

"Simple Java Edition"

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;    }}

Leetcode Search in rotated Sorted Array II problem Solving report