https://leetcode.com/problems/single-element-in-a-sorted-array/#/description
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/100754/Java-Binary-Search-short-(7l)-O(log(n))-w-explanations
Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.
Example 1:
Input: [1,1,2,3,3,4,4,8,8] Output: 2
Example 2:
Input: [3,3,7,7,10,11,11] Output: 10
Note: Your solution should run in O(log n) time and O(1) space.
从数组递增有序和O(log n)时间复杂度推断，题目可以采用二分查找求解。
def singleNonDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
lo, hi = 0, len(nums) - 1
while lo < hi:
mi = (lo + hi) >> 1
if nums[mi] == nums[mi - 1]:
if (mi - 1) & 1:
hi = mi - 2
else:
lo = mi + 1
elif nums[mi] == nums[mi + 1]:
if (mi + 1) & 1:
lo = mi + 2
else:
hi = mi - 1
else:
return nums[mi]
return nums[lo]https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/100754/Java-Binary-Search-short-(7l)-O(log(n))-w-explanations
This method seems to be a bit simpler to understand, since it doesn't start with the left half and stays a little bit closer to the conventional solutions.
public static int singleNonDuplicate(int[] nums) {
int start = 0, end = nums.length - 1;
while (start < end) {
// We want the first element of the middle pair,
// which should be at an even index if the left part is sorted.
// Example:
// Index: 0 1 2 3 4 5 6
// Array: 1 1 3 3 4 8 8
// ^
int mid = (start + end) / 2;
if (mid % 2 == 1) mid--;
// We didn't find a pair. The single element must be on the left.
// (pipes mean start & end)
// Example: |0 1 1 3 3 6 6|
// ^ ^
// Next: |0 1 1|3 3 6 6
if (nums[mid] != nums[mid + 1]) end = mid;
// We found a pair. The single element must be on the right.
// Example: |1 1 3 3 5 6 6|
// ^ ^
// Next: 1 1 3 3|5 6 6|
else start = mid + 2;
}
// 'start' should always be at the beginning of a pair.
// When 'start > end', start must be the single element.
return nums[start];
}
public int singleNonDuplicate(int[] nums) {
int low = 0;
int high = nums.length-1;
int low = 0;
int high = nums.length-1;
while(low < high) {
int mid = low + (high - low)/2;
if(nums[mid] != nums[mid+1] && nums[mid] != nums[mid-1])
return nums[mid];
else if(nums[mid] == nums[mid+1] && mid % 2 == 0)
low = mid+1;
else if(nums[mid] == nums[mid-1] && mid % 2 == 1)
low = mid+1;
else
high = mid-1;
}
return nums[low];
}
My solution using binary search. lo and hi are not regular index, but the pair index here. Basically you want to find the first even-index number not followed by the same number.
public int singleNonDuplicate(int[] nums) {
// binary search
int n=nums.length, lo=0, hi=n/2;
while (lo < hi) {
int m = (lo + hi) / 2;
if (nums[2*m]!=nums[2*m+1]) hi = m;
else lo = m+1;
}
return nums[2*lo];
}
Longer than writing my own binary search, but I wanted to see how this would look. I let
Collections.binarySearch search in a special List object which returns -1 for indexes before the single element, 0 for the index of the single element, and 1 for indexes after the single element. So I run binary search for 0. Since my special List object computes its entries only on the fly when requested, this solution takes O(log n) time and O(1) space. public int singleNonDuplicate(int[] nums) {
List list = new AbstractList<Integer>() {
public int size() {
return nums.length;
}
public Integer get(int index) {
if ((index ^ 1) < size() && nums[index] == nums[index ^ 1])
return -1;
if (index == 0 || index % 2 == 0 && nums[index - 1] != nums[index])
return 0;
return 1;
}
};
return nums[Collections.binarySearch(list, 0)];
}
A variation, the helper
isOff tells whether the index is in the part that's "off" (the single element and everything after it): public int singleNonDuplicate(int[] nums) {
List list = new AbstractList<Integer>() {
public int size() {
return nums.length;
}
public Integer get(int index) {
return isOff(index) + isOff(index - 1);
}
int isOff(int i) {
return i == size() - 1 || i >= 0 && nums[i] != nums[i ^ 1] ? 1 : 0;
}
};
return nums[Collections.binarySearch(list, 1)];
}