例1【二分查找】
已知有序的序列 int[] a,
整数 x,要求找到一个刚好比x稍微大一点的元素位置
public class Main {
//区间左闭右开
static int f(int []a,int low,int high,int x){
if(high-low==1){
if(a[low]>x)
return low;
return high;
}
int mid = (low + high) / 2;
if(x=a[a.length-1])//不存在比x稍大的数字
return -1;
return f(a,0,a.length,x);
}
public static void main(String[] args) {
int []a={1,5,11,24,25,32,32,33,49,66,68,69,70,75,88,102,114,119,120,135,146,221,294,300,302,305,333};
System.out.println(f(a,32));
}
}
例2【最大连续部分和】
数组中整数有正有负
求连续一子段,使得和最大化
2,4,-7,5,2,-1,2,-4,3
最大连续段:5,2,-1,2
最大和为8
思路分析:利用分治思想,分成三个部分,前一半和后一半的最大连续和,以及中间往两边的最大值
public class Main {
//区间左闭右开
static int f(int []a,int low,int high){
if(high-low==1){
if(a[low]>0)//若a[low]<0 ,可以不取值
return a[low];
return 0;
}
int mid = (low + high)/2;
int left = f(a,low,mid);//左半部分寻找最大连续部分和
int right = f(a,mid,high);//右半部分寻找最大连续部分和
int mid1 = 0;
int sum = 0;
for(int i=mid-1;i>=low;i--){
sum = sum + a[i];
if(sum>mid1) mid1=sum;
}
int mid2 = 0;
sum = 0;
for(int i=mid;imid2)
mid2 = sum;
}
int max = mid1 + mid2;//中间向两边扩展的最大连续部分和
if(max
例3【大数乘法】
multi("5935849584045839123456789","4595805849258430535")
public class Main {
static String add(String a,String b){//大数相加
if(a.length()<=8 && b.length()<=8)
return Integer.parseInt(a) + Integer.parseInt(b) + "";
String a1 = "0";
String a2 = a;
if(a.length()>8){//将a的低8位分离出
a1 = a.substring(0,a.length()-8);
a2 = a.substring(a.length()-8);
}
String b1 = "0";
String b2 = b;
if(b.length()>8){//将a的低8位分离出
b1 = b.substring(0,b.length()-8);
b2 = b.substring(b.length()-8);
}
String t = add(a2,b2);
if(t.length()>8)
return add(add(a1,b1),"1") + t.substring(1);
if(t.length()<8){
return add(a1,b1) + zero(8-t.length())+ t;
}
return add(a1,b1) + t;
}
static String zero(int x){
if(x==0) return "";
if(x==1) return "0";
return zero(x/2) + zero(x/2) + zero(x%2);
}
static String multi(String a,String b){//大数相乘
if(a.length()<=4 && b.length()<=4)
return Integer.parseInt(a) * Integer.parseInt(b) + "";
if(a.length()>4){
int mid = a.length()/2;
String s1 = a.substring(0,mid);
String s2 = a.substring(mid);
return add(multi(s1,b) + zero(s2.length()),multi(s2,b));
}
return multi(b,a);
}
public static void main(String[] args) {
System.out.println(multi("12345678900987654321666","1234567890123456789555"));
System.out.println(new BigInteger("12345678900987654321666").multiply(new BigInteger("1234567890123456789555")));//验证
}
}