Leetcode 81 Search in Rotated Sorted Array II
題目來源:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/description/
Follow up for “Search in Rotated Sorted Array”:
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7
might become 4 5 6 7 0 1 2
).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
題目分析:本題與 Search in Rotated Sorted Array唯一的區別是數組中元素可以重複,但是因爲這個條件,會出現一些比較複雜的情況,影響算法的時間複雜度,原來我們依靠中間元素和邊緣元素的大小關係,來判斷哪些部分仍然是有序的,現在因爲重複的出現,如果我們遇到中間和邊緣相等的情況,我們就丟失了哪邊有序的信息,因爲哪邊都有可能是有序的結果,不知道往哪邊走。解決的辦法只能是對邊緣移動一步,直到邊緣和中間不在相等或者相遇.
舉例:假如原數組爲1111115,翻轉後 1151111
.left=0,mid=3,target=5,A[left] == A[mid]爲true,有兩種可能性:
(1)A[left]到A[right]之間全部都是1
(2)A[left]到A[right]之間存在不同的數字,可能包括target,無法確定屬於那種情況。
若翻後數組爲11111111...115
.,或者爲全1,有最壞的時間複雜度O(n);
public boolean search(int[] nums, int target) {
int len=nums.length;
int left=0;
int right=len-1;
while(left<=right){
int mid=(left+right)/2;
if(nums[mid]==target)
return true;
if(nums[mid]<nums[left]){
if(nums[mid]<target&&nums[right]>=target){
left=mid+1;
}
else{
right=mid-1;
}
}
else if(nums[mid]>nums[left]){
if(nums[mid]>target&&nums[left]<=target)
right=mid-1;
else{
left=mid+1;
}
}
else{
left++;
}
}
return false;
}