位运算

# 7. 整数反转 (opens new window)

class Solution {
    public int reverse(int x) {
        int ret = 0;
        while(x!=0){
            if(ret>Integer.MAX_VALUE/10 || ret<Integer.MIN_VALUE/10)
                return 0;
            ret = ret*10 + x%10;
            x /= 10;
        }
        return ret;
    }
}

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# 9. 回文数 (opens new window)

class Solution {
    public boolean isPalindrome(int x) {
        // 负数不是回文，最后一位数字为0则第一位数字也得为0
        if(x<0 || (x%10==0&&x!=0))
            return false;
        int rev = 0;
        while(x>rev){
            rev = rev*10 + x%10;
            x /= 10;
        }
        // 数字长度分为 奇数或偶数
        return x==rev || x==rev/10;
    }
}

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# 136. 只出现一次的数字 (opens new window)

方法一：位运算

class Solution {
    public int singleNumber(int[] nums) {
        int ret = 0;
        for(int num:nums){
            ret ^= num;
        }
        return ret;
    }
}

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方法二：API

public class Solution {
    public int hammingWeight(int n) {
        return Integer.bitCount(n);
    }
}

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# 191. 位1的个数 (opens new window)

方法一：逐位判断

class Solution {
    public int hammingWeight(int n) {
        int res = 0;
        while(n!=0){
            res += n&1;
            n >>>= 1;
        }
        return res;
    }
}

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方法二：n&(n-1)

最右边的1变为0；

class Solution {
    public int hammingWeight(int n) {
        int res = 0;
        while(n!=0){
            res++;
            n &= (n-1);
        }
        return res;
    }
}

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← 优先级队列位运算总结→