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Comments (4)

PisecesPeng avatar PisecesPeng commented on June 24, 2024

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.

PisecesPeng avatar PisecesPeng commented on June 24, 2024

LeetCode题解

解题思路

  • 当检查第i位时,我们可以让n2^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.

PisecesPeng avatar PisecesPeng commented on June 24, 2024

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.

PisecesPeng avatar PisecesPeng commented on June 24, 2024

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.

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