Question

我正在练习CCC 2017 J4问题，但遇到了时间限制错误。我的代码超出了问题的时间限制（1秒）。问题链接如下（见问题J4）： CCC 2017

我还附上了问题的代码：

    public static void main(String[] args) {

        int starttime = new Scanner(System.in).nextInt();
        int hr = 12, hr1 = 0, hr2 = 0;
        int mn = 1, mn1 = 0, mn2 = 0;
        int counter = 0;
        int diff1 = 0, diff2 = 0, diff3 = 0;

        for (int i = 0; i < starttime; i++, mn++) {

            if (mn >= 60) {
                mn -= 60;
                if (hr + 1 < 13) {
                    hr += 1;
                } else {
                    hr = (hr + 1) % 12;
                }
            }
            mn1 = mn % 10;
            mn2 = mn / 10 % 10;
            hr1 = hr % 10;
            hr2 = hr / 10 % 10;
            diff1 = mn1 - mn2;
            diff2 = hr1 - hr2;
            diff3 = mn2 - hr1;

            if (hr2 == 0) {
                if (diff1 == diff3) {
                    counter += 1;
                }
            } else if (diff1 == diff2 && diff2 == diff3) {
                counter += 1;
            }
        }
        System.out.println(counter);
    }

Answer 1

通常有两种方法可以提高代码的速度。如果您不熟悉Big O表示法，请花一点时间阅读它。

现在你的代码是O（n），这完全没问题，但是如果你能找出一个降低它的技巧，（即O（log n））会提高运行速度。
如果您无法找到降低Big O复杂度的方法，那么另一个选择是减少输入/步数。因此，如果您可以找到某种技巧来在流程早期抛出输入，那么您的代码运行速度会稍快。

Answer 2

有一个更好的算法是O（1）。你只需找到一个聪明的解决方案就可以了。

这是一个javascript代码段：

＆＃13;

let seriesInMinutesFrom12oClock = [34]; // You'll need an linked list in java (or any type of list)
// So 12:34 is 34min and 1:23 and 83min ...
// Lets start with 12:34 already in the array since it's the only serie in the 12:00-12:59 timespan. Now we
// don't even have to deal with the fact that 12 really means zero.

for(let hours = 1; hours < 12; hours++) { // All the hours from 1 to 11 (no need to do 12, it's already done)
    for(let minutes = 0; minutes < 60; minutes++) { // 0 to 59

        let hoursTens = parseInt(hours / 10, 10); // Java equivalent of " int hoursTens = (int) (hours/10); "
        let hoursDigits = hours % 10;
        let minutesTens = parseInt(minutes / 10, 10);
        let minutesDigits = minutes % 10;

        let isArithmeticSerie = false;

        if (hoursTens == 0) {
            // Example (1:23)
            // 3 - 2 == 2 - 1
            isArithmeticSerie = (minutesDigits - minutesTens) == (minutesTens - hoursDigits);

        } else {
            // Example (12:34)
            // (4 - 3) == (3 - 2) && (3 - 2) == (2 - 1)
            isArithmeticSerie = (minutesDigits - minutesTens) == (minutesTens - hoursDigits) && (minutesTens - hoursDigits) == (hoursDigits - hoursTens);
        }

        if (isArithmeticSerie)
            seriesInMinutesFrom12oClock.push( // Add to list of series we found
                hoursTens * 600 +
                hoursDigits * 60 +
                minutesTens * 10 +
                minutesDigits
            );
    }
}

// Now you have your list for one cycle of 12hrs... You do that at the beginning before you ask for the input.

// Your program can start reading user input
let numberOfMinutesToObserve = 180; // minutes (read from System.in in java)


let numberOf12HrsCycles = numberOfMinutesToObserve / (12 * 60);

let numberOfFullCycles = parseInt(numberOf12HrsCycles, 10);

let leftOverMinutes = (numberOf12HrsCycles - numberOfFullCycles) * 12 * 60; // Last incomplete cycle (in minutes)

// The number of series is the number of full cycles times the number of series by cycles
let numberOfSeries = seriesInMinutesFrom12oClock.length * numberOfFullCycles;

// Add to it the number of arithmetic series that are in the last (incomplete) cycle of 12hrs
numberOfSeries += seriesInMinutesFrom12oClock.filter(mins => mins <= leftOverMinutes).length;
// The above line means: "find the number of series found before the end of the last cycle".

console.log(numberOfSeries);

＆＃13;

如何使我的Java程序更快？

2 个答案: