LeetCode复习 位运算系列

剑指 Offer 64 求1+2+…+n

int sumNum = 0;
    public int sumNums(int n) {
        //短路效应 只要n小于等于1，sumNums(n-1) > 0就不会执行
        boolean x = (n > 1) && sumNums(n-1) > 0;
        return sumNum += n;
    }

剑指 Offer 56 - II 数组中数字出现的次数 II

public int singleNumber2(int[] nums) {
        int[] count = new int[32];
        for (int num : nums) {
            for (int i = 0; i < 32; i++) {
                count[i] += num & 1;
                num >>= 1;
            }
        }
        int res = 0;
        for (int i = 31; i >= 0; i--) {
            //先左移是为了保留计算结果
            res <<= 1;
            res |= count[i] % 3;
        }
        return res;
    }

剑指 Offer 56 - I 数组中数字出现的次数

public int[] singleNumbers(int[] nums) {
        int res = 0;
        for (int num : nums) {
            res ^= num;
        }
        int mask = 1;
        for (int i = 0; i < 32; i++) {
            if ((res&mask) == mask) {
                break;
            }
            mask <<= 1;
        }
        int res1 = 0, res2 = 0;
        for (int num : nums) {
            if ((num & mask) == mask) {
                res1 ^= num;
            } else {
                res2 ^= num;
            }
        }
        return new int[]{res1,res2};
    }

剑指 Offer 65 不用加减乘除做加法

//用异或和与运算实现
 public int add(int a, int b) {
        //不为0表示还有进位
        while (b != 0) {
            int temp = b;
            //进位
            b = (a & b) << 1;
            //没有进位的和
            a = (a ^ temp);
        }
        return a;
    }

剑指 Offer 53 - II 0～n-1中缺失的数字

public int missingNumber(int[] nums) {
        int res = 0;
        int length = nums.length;
        for (int i = 0; i < length; i++) {
            res ^= nums[i] ^ i;
        }
        return res ^ length;
    }

剑指 Offer 15 二进制中1的个数

public int hammingWeight(int n) {
        int mask = 1, res = 0;
        for (int i = 0; i < 32; i++) {
            if ((mask & n) == mask) {
                res++;
            }
            mask <<= 1;
        }
        return res;
    }

汉明距离

public int hammingDistance(int x, int y) {
        int a = x ^ y;
        int mask = 1;
        int res = 0;
        for (int i = 0; i < 32; i++) {
            if ((a & mask) == mask) {
                res++;
            }
            a <<= 1;
        }
        return res;
    }

比特位计数

public int[] countBits(int n) {
        int[] dp = new int[n+1];
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            if (i % 2 == 1) {
                dp[i] = dp[i - 1] + 1;
            } else {
                dp[i] = dp[i/2];
            }
        }
        return dp;
    }

只出现一次的数字

 public int singleNumber(int[] nums) {
        int res = nums[0];
        for (int i = 1; i < nums.length; i++) {
            res ^= nums[i];
        }
        return res;
    }

二进制求和

 public String addBinary(String a, String b) {
        int length = a.length();
        int length1 = b.length();
        char[] chars = a.toCharArray();
        char[] chars1 = b.toCharArray();
        int i = length - 1, j = length1 - 1;
        StringBuffer sb = new StringBuffer();
        int carry = 0;
        int curSum = 0;
        while (i >= 0 || j >= 0 || carry > 0) {
            int c = i >= 0 ? chars[i] - '0' : 0;
            int d = j >= 0 ? chars1[j] - '0' : 0;
            curSum = c + d + carry;
            //是否有进位
            carry = curSum >= 2 ? 1 : 0;
            //有进位就减2
            curSum = carry == 1 ? curSum -= 2 : curSum;
            sb.append(curSum);
            i--;
            j--;
        }
        return sb.reverse().toString();
    }