算法 -- 倍率实验

今天学习到一个有意思的算法结论

如果 T(N) ~ aN^b * logN，那么 T(2N)/T(N) ~ 2^b。

实验对象：ThreeSum算法

时间复杂度为O(n^3)

public interface ThreeSum {
    int count(int[] nums);
}

public class ThreeSumSlow implements ThreeSum {
    @Override
    public int count(int[] nums) {
        int N = nums.length;
        int cnt = 0;
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) {
                for (int k = j + 1; k < N; k++) {
                    if (nums[i] + nums[j] + nums[k] == 0) {
                        cnt++;
                    }
                }
            }
        }
        return cnt;
    }
}

时间测试方法

public class StopWatch {

    private static long start;


    public static void start() {
        start = System.currentTimeMillis();
    }


    public static double elapsedTime() {
        long now = System.currentTimeMillis();
        return (now - start) / 1000.0;
    }
}

public class RatioTest {
    
    public static void main(String[] args) {
        int N = 500;
        int loopTimes = 7; //循环次数
        double preTime = -1; //每次循环的起始时间
        while (loopTimes-- > 0) {
            int[] nums = new int[N];
            StopWatch.start();
            ThreeSum threeSum = new ThreeSumSlow();
            int cnt = threeSum.count(nums);
            System.out.println(cnt);
            double elapsedTime = StopWatch.elapsedTime(); //每次循环的结束时间
            double ratio = preTime == -1 ? 0 : elapsedTime / preTime;
            System.out.println(N + "  " + elapsedTime + "  " + ratio);//倍率
            preTime = elapsedTime;
            N *= 2;
        }
    }
}

实验结果

确实如同开始的结论，2^3 == > a * O(n^3) * logn