題目：

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 nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.Example: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.

方案一：找規(guī)律，時間復雜度為O(1).

public int bulbSwitch(int n) {return (int)Math.sqrt(n);}

方案二：笨法子，按題目敘述逐行實現，時間復雜度為O(N2)，系統(tǒng)超時.

public int bulbSwitch(int n) {int[] arrBulbStatus = new int[n];for (int i = 1; i <= n; i++) { // on the i roundfor (int j = i-1; j < arrBulbStatus.length; j += i) { // toggle every i bulbif (arrBulbStatus[j] == 1) {arrBulbStatus[j] = 0;} else {arrBulbStatus[j] = 1;}}}int res = 0;for (int i = 0; i < arrBulbStatus.length; i++) { // count how many bulbs are on.if (arrBulbStatus[i] == 1) {res++;}}return res;}

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轉載于:https://www.cnblogs.com/lasclocker/p/5154118.html

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