Search in Rotated Sorted Array II LeetCode Java
阿新 • • 發佈:2018-06-11
ret RM In arr duplicate true rst array AS
描述
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.
分析
允許重復元素,則上一題中如果 A[m]>=A[l], 那麽 [l,m] 為遞增序列的假設就不能成立了,比
如 [1,3,1,1,1]。
如果 A[m]>=A[l] 不能確定遞增,那就把它拆分成兩個條件:
? 若 A[m]>A[l],則區間 [l,m] 一定遞增
? 若 A[m]==A[l] 確定不了,那就 l++,往下看一步即可。
二分查找
代碼
1 // LeetCode, Search in Rotated Sorted Array II 2 // 時間復雜度 O(log n),空間復雜度 O(1) 3 class Solution { 4 public static boolean search(int A[], int n, int target) { 5 int first = 0, last = n; 6 while (first != last) { 7 intmid = (first + last) / 2; 8 if (A[mid] == target) 9 return true; 10 if (A[first] < A[mid]) { 11 if (A[first] <= target && target < A[mid]) 12 last = mid;13 else 14 first = mid + 1; 15 } else if (A[first] > A[mid]) { 16 if (A[mid] <= target && target <= A[last-1]) 17 first = mid + 1; 18 else 19 last = mid; 20 } else 21 //skip duplicate one, A[start] == A[mid] 22 first++; 23 } 24 return false; 25 } 26 }
Search in Rotated Sorted Array II LeetCode Java