數(shù)組拷貝，二分法查找

Arrays.copyOfRange(int[] arr,int開(kāi)始下標(biāo)，int結(jié)束下標(biāo))//左閉右開(kāi)
System.arraycopy（int[] source,int開(kāi)始下標(biāo)，int[] dest,int開(kāi)始下標(biāo)，int拷貝多少個(gè)）

這里是手動(dòng)拷貝：
public static int[] copyOf(int[] original, int newLength) {
    int[] dest = new int[newLength];
    int length = original.length <= newLength ? original.length : newLength;
    for (int i = 0; i < length; i++) {
        dest[i] = original[i];
    }

    return dest;
}

mid作為分水嶺，  不斷地縮小范圍
// 前提需要數(shù)組是有序的
public static int binarySearch(int[] a, int v) {
    int left = 0;
    int right = a.length;

    while (left < right) {
        int mid = (left + right) / 2;
        if (v == a[mid]) {
            return mid;
        } else if (v < a[mid]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }

    return -1;
}

public static int binarySearch3(int[] a, int v) {
    int left = 0;
    int right = a.length - 1;

    while (left <= right) {
        int mid = (left + right) / 2;
        if (v == a[mid]) {
            return mid;
        } else if (v < a[mid]) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }

    return -1;
}