给每个数循环做x & (x - 1)并计数就是它二进制数中1的个数。
class Solution {
public:
vector<int> countBits(int num) {
vector<int> ret;
for (int i = 0; i <= num; ++i) {
ret.push_back(count(i));
}
return ret;
}
int count(int x) {
int ret = 0;
while (x) {
x = x & (x - 1);
++ret;
}
return ret;
}
};
分析 给每个数循环做x & (x - 1)并计数就是它二进制数中1的个数。
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本文标题：LeetCode 338-Counting Bits
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