Question

有人可以帮助我解决以下代码的每个部分的时间复杂性，包括for循环和if循环以及所有这些。

public class GaleShapley {
    private int N, engagedCount;
    private int[][] menPref;
    private int[][] womenPref;
    private int[] men;
    private int[] women;
    private int[] womenPartner;
    private boolean[] menEngaged;

    /** Constructor **/
    public GaleShapley(int[] m, int[] w, int[][] mp, int[][] wp) {
        N = mp.length;
        engagedCount = 0;
        men = m;
        women = w;
        menPref = mp;
        womenPref = wp;
        menEngaged = new boolean[N];
        womenPartner = new int[N];
        calcMatches();
    }

    /** function to calculate all matches **/
    private void calcMatches() {

        while (engagedCount < N) {
            int free;
            for (free = 0; free < N; free++)
                if (!menEngaged[free])
                    break;

            for (int i = 0; i < N && !menEngaged[free]; i++) {
                int index = womenIndexOf(menPref[free][i]);
                if (womenPartner[index] == 0) {
                    womenPartner[index] = men[free];
                    menEngaged[free] = true;
                    engagedCount++;
                } else {
                    int currentPartner = womenPartner[index];
                    if (morePreference(currentPartner, men[free], index)) {
                        womenPartner[index] = men[free];
                        menEngaged[free] = true;
                        menEngaged[menIndexOf(currentPartner)] = false;
                    }
                }
            }
        }
        printCouples();
    }

    /** function to check if women prefers new partner over old assigned partner **/
    private boolean morePreference(int curPartner, int newPartner, int index) {
        for (int i = 0; i < N; i++) {
            if (womenPref[index][i] == newPartner)
                return true;
            if (womenPref[index][i] == curPartner)
                return false;
        }
        return false;
    }

    /** get men index **/
    private int menIndexOf(int str) {
        for (int i = 0; i < N; i++)
            if (men[i] == str)
                return i;
        return -1;
    }

    /** get women index **/
    private int womenIndexOf(int str) {
        for (int i = 0; i < N; i++)
            if (women[i] == str)
                return i;
        return -1;
    }

    /** print couples **/
    public void printCouples() {

        System.out.println("Couples are : ");
        for (int i = 0; i < N; i++) {
            System.out.println(women[i] + " " + womenPartner[i]);
        }
    }

    /** main function **/
    public static void main(String[] args) {
        System.out.println("Gale Shapley Marriage Algorithm\n");
        /** list of men **/
        int[] m = { 1, 2, 3, 4, 5 };
        /** list of women **/
        int[] w = { 1, 2, 3, 4, 5 };

        /** men preference **/
        int[][] mp = { { 2, 4, 1, 3, 5 }, { 3, 1, 4, 2, 5 }, { 2, 3, 1, 4, 5 },
                { 4, 1, 3, 2, 5 }, { 2, 1, 4, 3, 5 } };
        /** women preference **/
        int[][] wp = { { 2, 1, 4, 3 }, { 4, 3, 1, 2 }, { 1, 4, 3, 2 },
                { 2, 1, 4, 3 }, { 2, 1, 4, 3 } };

        GaleShapley gs = new GaleShapley(m, w, mp, wp);
    }
}

我假设时间复杂度是O（n ^ 2）。需要有人确认一样。

Answer 1

你的功能以while循环（从0到N）开始。

在此循环中，您有2个for循环（从0到N）。

在第二个for循环中，您调用方法morePreference，menIndexOf和womenIndexOf。这些方法中的每一个都包含一个for循环（从0到N）。

因此，您的算法的复杂性为O(N^3)

使用BigO表示法编写此代码的时间复杂度

1 个答案: