Comments (4)
解题思路
- 将数字转换为数字, 再遍历判断'1'存在的个数.
代码
public static int func(int n) {
int count = 0;
for (char c : String.valueOf(n).toCharArray())
if ('1' == c) count++;
return count;
}
from pisecespeng.record.me.
LeetCode题解
解题思路
- 当检查第
i
位时,我们可以让n
与2^i
进行与运算, - 当且仅当
n
的第i
位为1
时, 运算结果不为0
.
代码
public static int func(int n) {
int ret = 0;
for (int i = 0; i < 32; i++) {
if ((n & (1 << i)) != 0) {
ret++;
}
}
return ret;
}
from pisecespeng.record.me.
LeetCode题解
解题思路
- 位运算优化
代码
public static int func(int n) {
int ret = 0;
while (n != 0) {
n &= n - 1;
ret++;
}
return ret;
}
from pisecespeng.record.me.
LeetCode题解
解题思路
- 位运算优化
代码
public static int func(int n) {
int ret = 0;
while (n != 0) {
n &= n - 1;
ret++;
}
return ret;
}
from pisecespeng.record.me.
Related Issues (20)
- 两数相加 HOT 2
- 买卖股票的最佳时机 HOT 3
- 买卖股票的最佳时期2 HOT 3
- 二叉树的层序遍历 HOT 2
- 二叉树的最大深度 HOT 2
- 二进制求和 HOT 4
- 从尾到头打印链表 HOT 2
- 删除排序数组中的重复项 HOT 2
- 删除链表的倒数第N个节点 HOT 3
- 加一 HOT 2
- 反转字符串 HOT 2
- 反转链表 HOT 3
- 只出现一次的数字 HOT 2
- 合并两个有序数组 HOT 3
- 合并两个有序链表 HOT 2
- 回文数 HOT 2
- 回文链表 HOT 3
- 复制带随机指针的链表 HOT 2
- 外观数列 HOT 3
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