Leetcode 137. Single Number II
方法1: 136的升級版。依舊是bit manipulation。我感覺這題目老難了。我現在看了都沒咋看懂,更別說面試時候自己寫出來了。時間複雜n,空間複雜1.
class Solution {
public int singleNumber(int[] nums) {
int seenOnce = 0, seenTwice = 0;
for (int num : nums) {
// first appearence:
// add num to seen_once
// don't add to seen_twice because of presence in seen_once
// second appearance:
// remove num from seen_once
// add num to seen_twice
// third appearance:
// don't add to seen_once because of presence in seen_twice
// remove num from seen_twice
System.out.println(~seenTwice) ;
seenOnce = ~seenTwice & (seenOnce ^ num);
System.out.println(~seenOnce) ;
seenTwice = ~seenOnce & (seenTwice ^ num);
}
return seenOnce;
}
}
總結:
- 無
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