数组拷贝，二分法查找

这里是手动拷贝：
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作为分水岭， 不断地缩小范围
// 前提需要数组是有序的
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 binarySearch2(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;
}
}