第75场双周赛

6033. 转换数字的最少位翻转次数 - 力扣（LeetCode） (leetcode-cn.com)

class Solution {
public:
    int minBitFlips(int start, int goal) {
        int bit = 1;
        uint32_t t = start, t1 = goal;
        int res = 0, i = 0;
        while (start != goal) {
            int x = (start >> i) & 1;
            int y = (goal >> i) & 1;
            if (x != y) {
                if (y == 0) {
                    start &= ~(1 << i);
                } else {
                    start |= (1 << i);
                }
                res ++;
            }
            i ++;
            
        }
        return res;
    }
};

6034. 数组的三角和 - 力扣（LeetCode） (leetcode-cn.com)

模拟队列

class Solution {
public:
    int triangularSum(vector<int>& nums) {
        queue<int> q;
        for (auto num: nums) q.push(num);
        int bunner = nums.size() - 1, i = 0;
        while (q.size() != 1) {
            int a = q.front();
            q.pop();
            int b = q.front();
            int sum = a + b;
            q.push(sum % 10);
            i ++;
            if (i == bunner) {
                q.pop();
                i = 0;
                bunner = q.size() - 1;
            }
        }
        return q.front();
    }
};

空间复杂度1做法

class Solution {
public:
    int triangularSum(vector<int>& nums) {
        int n = nums.size();
        for (int i = n - 1; i > 0; i --) 
            for (int j = 0; j < i; j ++)
                nums[j] = (nums[j] + nums[j + 1]) % 10;
        return nums[0];
    }
};

6035. 选择建筑的方案数 - 力扣（LeetCode） (leetcode-cn.com)

class Solution {
public:
    long long numberOfWays(string s) {
        long long n1 = 0, n01 = 0, n10 = 0, n0 = 0;
        long long res = 0;
        for (auto &c: s) {
            if (c == '1') {
                n01 += n0;
                n1 ++;
                res += n10;
            } else {
                n10 += n1;
                n0 ++;
                res += n01;
            }
        }
        return res;
    }
};

前缀和

class Solution {
public:
    long long numberOfWays(string s) {
        int n = s.size();
        vector<int> count(n, 0);
        int zeroCount = 0, oneCount = 0;
        for (int i = n - 1; i >= 0; i --) {
            if (s[i] == '0') {
                count[i] = oneCount;
            } else {
                count[i] = zeroCount;
            }
            zeroCount += s[i] == '0' ? 1 : 0;
            oneCount += s[i] == '1' ? 1 : 0;
        }
        zeroCount = oneCount = 0;
        long long res = 0;
        for (int i = 0; i < n; i ++) {
            if (s[i] == '0') {
                count[i] *= oneCount;
            } else {
                count[i] *= zeroCount;
            }
            zeroCount += s[i] == '0' ? 1 : 0;
            oneCount += s[i] == '1' ? 1 : 0;
            res += count[i];
        }
        return res;
    }
};