你对MySQL的慢查询优化有了解吗
没有了解
https://blog.csdn.net/weixin_45764765/article/details/124112232
约瑟夫环剩余问题
模拟:有特殊用例超时通过不了
class Solution {
public:
int LastRemaining_Solution(int n, int m) {
vector<int>children(n);
int count = n;
int dex = 0;
int step = 0;
while(count > 0){
if(dex >= n){
dex = 0;
}
if(children[dex] == -1){
dex++;
continue;
}
if(step == m-1){
children[dex] = -1;
count--;
step = -1;
}
dex++;
step++;
}
return dex-1;
}
};
数学加递归
class Solution {
int f(int n, int m) {
if (n == 1) {
return 0;
}
int x = f(n - 1, m);
return (m + x) % n;
}
public:
int lastRemaining(int n, int m) {
return f(n, m);
}
};
作者:LeetCode-Solution
链接:https://leetcode.cn/problems/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-lcof/solution/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-by-lee/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
class Solution {
public:
int lastRemaining(int n, int m) {
int f = 0;
for (int i = 2; i != n + 1; ++i) {
f = (m + f) % i;
}
return f;
}
};
作者:LeetCode-Solution
链接:https://leetcode.cn/problems/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-lcof/solution/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-by-lee/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。