資訊專欄INFORMATION COLUMN
摘要：復(fù)雜度思路每次通過(guò)二分法找到一個(gè)值之后，搜索整個(gè)數(shù)組，觀察小于等于這個(gè)數(shù)的個(gè)數(shù)。考慮，小于這個(gè)位置的數(shù)的個(gè)數(shù)應(yīng)該是小于等于這個(gè)位置的。要做的就是像找中的環(huán)一樣，考慮重復(fù)的點(diǎn)在哪里?？紤]用快慢指針。代碼把一個(gè)指針放回到開頭的地方
LeetCode[287] Find the Duplicate Number
Binary SearchGiven an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
復(fù)雜度
O(NlgN), O(1)
思路
每次通過(guò)二分法找到一個(gè)值之后，搜索整個(gè)數(shù)組，觀察小于等于這個(gè)數(shù)的個(gè)數(shù)。
考慮[1,2,3,2]，小于這個(gè)位置的數(shù)的個(gè)數(shù)應(yīng)該是小于等于這個(gè)位置的。如果cnt > mid,說(shuō)明在小于這個(gè)位置的數(shù)的范圍內(nèi)，存在duplicate，right = mid - 1;
代碼
public int findDuplicate(int[] nums) {
int left = 0, right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left) / 2;
int cnt = 0;
for(int i = 0; i < nums.length; i ++) {
if(nums[i] <= mid) cnt ++;
}
if(cnt > mid) right = mid - 1;
else left = mid + 1;
}
return left;
}
LinkedList判斷循環(huán)
復(fù)雜度
O(N), O(1)
思路
考慮一定會(huì)有duplicate出現(xiàn)，說(shuō)明，數(shù)組里面的值組成了一個(gè)環(huán)，如果考慮像linkedlist那樣。
要做的就是像找linkedlist中的環(huán)一樣，考慮重復(fù)的點(diǎn)在哪里。
考慮用快慢指針。
代碼
public int findDuplicate(int[] nums) {
if(nums.length > 1) {
int slow = 0;
int fast = 0;
while(true) {
slow = nums[slow];
fast = nums[nums[fast]];
if(slow == fast) {
// case like [1,2,3,1]
// 把一個(gè)指針放回到開頭的地方
slow = 0;
while(slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
}
}
return -1;
}
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Problem Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate nu...
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