Link: https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/
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.
Time: O(n) //because of linear search due to duplicates. Space: O(1)public class Solution {
public boolean search(int[] A, int target) {
int l = 0;
int r = A.length - 1;
while(l <= r){
int m = (l+r)/2;
if(A[m] == target) return true;
else if(A[m] > A[l]){
//should add to A[l] <= target, not target < A[m] because if target == A[m], we should not decrease high
if(A[l] <= target && target < A[m]){
r = m - 1;
}
else{
l = m + 1;
}
}
else if (A[m] < A[l]){
if(A[m] < target && target <=A[r]){
l = m + 1;
}
else{
r = m - 1;
}
}
else{//A[m] == A[l]
l++;
}
}
return false;
}
}
Note: 11.3.2014
一遍沒過。記得判斷A[m] 和A[l]的大小。
while(l<=r), not l < r. e.g. When A = {1}, target = 1, if while(l <r) then we will return false.