位运算

剑指 Offer II 004. 只出现一次的数字

刷穿剑指offer-Day02-整数II 004.只出现一次的数字 位运算讲解 - 只出现一次的数字

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int ret = 0;
        for (int i = 0; i < 32; i ++) { 
            int cnt = 0;
            for (auto num: nums) {
                cnt += num >> i & 1;
            }
            if (cnt % 3 != 0) {
                ret |= 1 << i;
            }
        }
        return ret;
    }
};

剑指 Offer II 005. 单词长度的最大乘积

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int> dict(n, 0);
        for (int i = 0; i < words.size(); i ++) {
            int t = 0;
            for (auto c: words[i]) {
                int bit = c - 'a';
                t |= (1 << bit);
            }
            dict[i] = t;
        }

        int ans = 0;
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < i; j ++) 
                if ((dict[i] & dict[j]) == 0) {
                    int temp = words[i].size() * words[j].size();
                    ans = max(ans, temp);
                }
        return ans;
    }   
};