1. ???? ??

[??]

?? ???? ??? n?? ??? ???? ???. (0?? ???? 0?? ??? 0???.) ).

(?? ??? ????: java ??)

[Code]

package swear2offer.array;


public class FeiBoNaQi {

    /**
     * 大家都知道斐波那契數(shù)列，現(xiàn)在要求輸入一個(gè)整數(shù) n，
     * 請(qǐng)你輸出斐波那契數(shù)列的第 n 項(xiàng)（從 0 開(kāi)始，第 0 項(xiàng)為 0）。
     * 0,1,1,2,3,5
     * n<=39
     * */
    public int Fibonacci(int n) {
        if (n == 0) return 0;

        if (n == 1 || n== 2) return 1;

        return Fibonacci(n-1) + Fibonacci(n-2);
    }

    /**
     * 非遞歸方式，遞歸的數(shù)據(jù)結(jié)構(gòu)使用的棧，那就是使用棧的方式
     * */
    public int NoRecursive(int n) {
        if (n>2) {
            int[] a = new int[n+1];
            a[0] = 0;
            a[1] = 1;
            a[2] = 1;

            for (int i=3; i<=n; i++) {
                a[i] = a[i-1] + a[i-2];
            }

            return a[n];
        } else {
            if (n == 0) return 0;
            else return 1;
        }
    }

    public static void main(String[] args) {
        System.out.println(new FeiBoNaQi().Fibonacci(39));
        System.out.println(new FeiBoNaQi().Fibonacci(39));
    }
}

2. ???? ??

[??]

21?? ?? ????? ???? ?? ?? ??? ? ? ????? ?? ? ????. ??? 2*n? ? ????? ??? 21? n?? ?? ?????? ??? ?? ?? ??? ? ??????

?? ?? n=3? ?? 2*3 ???? ???? 3?? ?? ??? ????.

??:

package swear2offer.array;

public class Rectangle {

    /**
     * f[0] = 0;
     * f[1] = 1;
     * f[2] = 2;
     * f[3] = 3;
     * f[4] = 5;
     * f[5] = 8;
     * f[n] = f[n-1] + f[n-2]
     * */

    public int RectCover(int target) {

        if (target < 4) return target;

        int[] f = new int[target + 1];
        int i;
        for (i=0; i<4; i++) f[i] = i;

        for (i=4; i<=target; i++) {
            f[i] = f[i-1] + f[i-2];
        }

        return f[target];
    }



    public static void main(String[] args) {
        System.out.println(new Rectangle().RectCover(5));
    }
}

[Thinking]

??? ???? ?? ??? ??? ??? ?? ????. , ??? ??? ?? ??? ???? ??? ???? ???? ? ? ???, ?? ??? ??? ??? ?? ?? ?? n? ??? ?? ???? ???? ????. n-1? ???? ? ?? ??? ??(f[n-2])?? ??(f[n-1])?? ??? ??? ?????.

(?? ??? ?? ??: java ??? ?? ? ??)

3.2?? 1

[??]

??? ???? 1? ??? ???? ?????. ??? 2? ??? ?????.

【??】

package swear2offer.array;

public class Binary {

    /**
     * 輸入一個(gè)整數(shù)，輸出該數(shù)二進(jìn)制表示中 1 的個(gè)數(shù)。其中負(fù)數(shù)用補(bǔ)碼表示。
     * */
    public int NumberOf1(int n) {
        int count;
        count = 0;

        while(n != 0) {
            n = n & (n-1);// 與操作就是二進(jìn)制的操作，適用原碼和補(bǔ)碼
            count ++;
        }

        return count;
    }
}

【??】

??? ?? ??: ?? ??? ??? ????, ??? ??? ?????. ??? ??: ?? ??? ???? +1.

??? 0? ?? ?? ? ??? ?? ? ??? 1???. ? ???? 1? ?? ??? ?? ???? ?? ?? 1? 0? ??, ?? 1 ?? ?? 0? 1? ???(?? ??? 1 ?? 0? ?? ??). ??? ?? ??? ??? ?? ????.

?: ??? 1100? ?? ????? ? ?? ??? ?? ??? 1???. 1? ?? ? ?? ??? 0? ??, ???? ? ?? 0? 1? ??, ? ?? 1? ??? ?? ???? ??? 1011? ???. 1? ? ??? ?? ???? ??? ??? ??? ? ? ????. 1? ???? ?? ??? ???. ??, ?? ??? 1? ? ??? AND ???? ?? ??? ?? ??? 1?? ???? ?? ??? 0? ???. ?? ??, 1100&1011=1000 ?, ???? 1? ? ?? ?? ??? AND ??? ???? ??? ?? ???? ?? 1? 0?? ????. ??? 1? ???? ????. ??? ?? ???? ?? ?? ?? ?????.

4. ?? ?? ????

[??]

double ??? ?? ??? ?? ??? int ??? ?? ??? ?????. ????? ????? ??? ????.
??? ??? ??? 0? ??? ?????

[Code]

package swear2offer.array;

public class Power {

    public double Power(double base, int exponent) {

        if (base == 0) return 0;
        if (exponent == 0) return 1;

        int count;
        boolean flag;
        double temp;

        count = 1;
        temp = base;
        flag = true;// 標(biāo)記正負(fù)

        if (exponent < 0){
            exponent = -exponent;
            flag = false;
        }

        while (count < exponent) {
            base *= temp;
            count ++;
        }

        if (flag) {
            return base;
        } else {
            return 1/base;
        }

    }

    public static void main(String[] args) {
        System.out.println(new Power().Power(2,-3));
    }

}

[Thinking]

? ??? ??? ??, ????? ?? ???? ???, ???? ??? ????

? ?? ???? ?????. ??? ??? ??? ??? ????

??, ??==0? ??==0? ??? ????

????? ??? ??? ? ??? ??? ? ????. ???? ?? ??? ???.

5. ??? ?? ?? ??? ??? ??? ?????

[Title]

?? ??? ???? ??? ?? ??? ???? ??? ?????. ?? ??? ??? ? ?? ??? ?? ?? ??? ??? ? ?? ??? ???, ??? ??, ??? ?? ??? ?? ??? ???? ?? ??? ????? ?????.

[??]

package swear2offer.array;

import java.util.Arrays;

public class OddEven {

    /**
     * 輸入一個(gè)整數(shù)數(shù)組，實(shí)現(xiàn)一個(gè)函數(shù)來(lái)調(diào)整該數(shù)組中數(shù)字的順序，
     * 使得所有的奇數(shù)位于數(shù)組的前半部分，所有的偶數(shù)位于數(shù)組的后半部分，
     * 并保證奇數(shù)和奇數(shù)，偶數(shù)和偶數(shù)之間的相對(duì)位置不變。
     *
     * 時(shí)空復(fù)雜度較高的算法：
     * 新建一個(gè)數(shù)組b，用來(lái)保存奇數(shù)，在重新變量一次，保存偶數(shù)
     * 時(shí)空復(fù)雜度0(n)
     * */
    public void reOrderArray1(int [] array) {
        int n,i,j;
        n = array.length;

        int[] b = new int[n];

        j = 0;// 用來(lái)保存數(shù)組B的下標(biāo)
        for (i=0; i<n; i++) {
            if (array[i]%2 != 0) {
                b[j] = array[i];
                j ++;
            }
        }
        for (i=0; i<n; i++) {
            if (array[i]%2 == 0){
                b[j] = array[i];
                j++;
            }
        }

        for (i=0; i<n; i++) {
            array[i] = b[i];
        }
    }

    /**
     * 采用的冒泡交換以及快速排序的思想：
     * 設(shè)定兩個(gè)游標(biāo)，游標(biāo)分別用來(lái)標(biāo)記奇數(shù)和偶數(shù)的下標(biāo)，然后交換二者
     * 注意交換二者是無(wú)法保證順序的，交換的ij之間還有進(jìn)行平移。
     * */
    public void reOrderArray(int [] array) {

        int n,i,j,temp,p;

        n = array.length;
        i = 0;
        j = 0;
        while (i<n && j<n) {
            // i標(biāo)記偶數(shù)下標(biāo)
            while (i<n) {
                if (array[i]%2 ==0){
                    break;
                } else {
                    i++;
                }
            }
            j = i;
            // j標(biāo)記奇數(shù)下標(biāo)
            while (j<n) {
                if (array[j]%2 !=0){
                    break;
                } else {
                    j++;
                }
            }
            if (i<n && j<n) {
                // 此時(shí)ij已經(jīng)在遇到的第一個(gè)偶數(shù)和奇數(shù)停下，把ij之間的內(nèi)容平移
                temp = array[j];
                for (p=j; p>i; p--) {
                    array[p] = array[p-1];
                }
                array[i] = temp;
                // 此時(shí)把i,j標(biāo)記到 未交換前的偶數(shù)位置的下一個(gè)
                i ++;
                j = i;
            }
        }
    }

    public static void main(String[] args) {
        int[] a = {1,4,6,3,2,5,8};
        int[] b = {2,4,6,1,3,5,7};
        new OddEven().reOrderArray(b);
        System.out.println(Arrays.toString(b));
    }
}

[??]

??? ? ??? ??? ??? ???? ?????. ? ?? ??? ? ???? ??? ???? ??? ????. . ??, ??? ? ?? ?? ?? ??? ? ??? ??? ?? ?????.

?? ?? ??: Java ????

? ??? Java ???? ???? ?? ?? ??(2)? ?? ?????. ??? ??? PHP ??? ????? ?? ?? ??? ?????!