319. Bulb Switcher解题报告

题目

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Given n = 3.

At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off].
So you should return 1, because there is only one bulb is on.

题目解析

用代码来解析这个题目过程:

class Solution {public:    int bulbSwitch(int n) {        if (n == 1) return n;        bool* p = new bool[n + 1];        for (int i = 0; i <= n; ++i) {            p[i] = true;        }        for (int i = 2; i <= n; ++i) {            for (int j = i; j <= n; j = j + i) {                p[j] = !p[j];            }        }        int result = 0;        for (int i = 1; i <= n; ++i) {            if (p[i] == true) result++;        }        delete []p;        return result;    }};

思路分析

用题目分析的代码，复杂度是O(n^2)，因此该代码提交会超时。对于每一个数字，无论n取多少，它的灯状态都是固定的，不变的，在这里可以列举前16个数字的灯状态：

T,F,F,T,F,F,F,F,T,F,F,F,F,F,F,T

可以发现，当数字为平方数时，它的状态永远是True，因此可以得到题目的解

AC代码

class Solution {public:    int bulbSwitch(int n) {        int result = 1;        while (true) {            if (result * result < n){ result++;}            else if (result * result == n) { return result;}            else {return result-1;}        }    }};