java学习之数组（二）

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紧接着上一篇数组博文的下一篇；上一篇有一个数三退一的游戏，在这里还有一个很能体现面向对象的解法，还运用了稍显高大上的双向环哦~

public class Count3Quit {    public static void main(String[] args) {        KidCircle kd = new KidCircle(500);        int countNum = 0;        Kid k = kc.first;        while(kc.count > 1) {            countNum++;            if(countNUm == 3) {                countNum = 0;                kc.delete(k);            }            k = k.right;        }        System.out.println(kc.first.id);    }   }class Kid {    int id;    Kid left;    Kid right;}class KidCircle {    int count = 0;    Kid first,last;    KidCircle(int n) {        for(int i = 0; i < n; i++) {            add();        }    }    void add() {        Kid k = new Kid();        k.id = count;        if(count <= 0) {            first = k;            last = k;            k.left = k;            k.right = k;        } else {            last.right = k;            k.left = last;            k.right = first;            first.left = k;            last = k;        }        count++;    }    void delete(Kid k) {        if(count <= 0) {            return;        } else if(count == 1) {            first = last = null;        } else {            k.left.right = k.right;            k.right.left = k.left;            if(k == first) {                first = k.right;            } else if(k == last) {                last = k.left;            }        }    }}接下来上一位数组里面一个不得不提的经典小程序，即二分法查找

public class TestSearch {
public static void main(String[] args) {
int a[] = {1,3,6,8,9,10,12,18,20,34};
int i = 12;
System.out.println(binarySearch(a,i));
}

public static int binarySearch(int[] a, int num) {    if(a.length == 0) {        return -1;    }    int startPos = 0;    int endPos = a.length - 1;    int m = (startPos + endPos) / 2;    while(startPos <= endPos) {        if(a[m] == num) {            return m;        }        if(a[m] > num) {            endPos = m - 1;        }        if(a[m] < num) {            startPos = m + 1;        }        m = (startPos + endPos) / 2;    }}

}
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