LeetCode——Single Number II
2014-06-24 13:32
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Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
给定一个整数数组,除一个外,其它每个元素都出现3次,找出那一个。
可以完全套用上一题的利用哈希表的解法。
另一种解法是:用变量ones 代表二进制1只出现一次的数位,twos 代表二进制1只出现2次的数位,threes
代表二进制1出现3次的数位。当ones和twos中的某一位同时为1时表示二进制1出现3次,此时需要清零。
public int singleNumber(int[] A) {
int ones = 0, twos = 0;
for (int i = 0; i < A.length; i++) {
twos = twos | (ones & A[i]);
ones = ones ^ A[i];
//把出现了3次的位置设置为0,取反
int threes = ~(ones & twos);
ones &= threes;
twos &= threes;
}
return ones;
}
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
给定一个整数数组,除一个外,其它每个元素都出现3次,找出那一个。
可以完全套用上一题的利用哈希表的解法。
另一种解法是:用变量ones 代表二进制1只出现一次的数位,twos 代表二进制1只出现2次的数位,threes
代表二进制1出现3次的数位。当ones和twos中的某一位同时为1时表示二进制1出现3次,此时需要清零。
public int singleNumber(int[] A) {
int ones = 0, twos = 0;
for (int i = 0; i < A.length; i++) {
twos = twos | (ones & A[i]);
ones = ones ^ A[i];
//把出现了3次的位置设置为0,取反
int threes = ~(ones & twos);
ones &= threes;
twos &= threes;
}
return ones;
}
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