各种算法题MARK

①、斐波那契数列

 1 static void method(){
 2         
 3         String a = "1",b="1";
 4         char[] aArr;
 5         char[] bArr;
 6         int num = 0;
 7         int i,j;
 8         int temp = 0;
 9         int k;
10         System.out.print(" "+a+" "+b);
11         for(i =2 ; i<=500;i++){
12             aArr = a.toCharArray();
13             bArr = b.toCharArray();
14             a = b;
15             k = bArr.length - aArr.length;
16             for(j = aArr.length-1;j>=0;j--){
17                 temp = (bArr[j+k]-'0')+(aArr[j]-'0') + num;
18                 num = 0;
19                 if(temp >=10){
20                     temp -=10;
21                     num = 1;
22                 }
23                 bArr[j+k] = (char) (temp + '0');
24             }
25             while(j+k>=0 &&num==1){
26                 temp = (bArr[j+k] - '0') + num;
27                 if(temp > 10){
28                     bArr[j+k] = (char) (bArr[j+k] + (temp - 10));
29                     num = 1;
30                 }else{
31                     num = 0;
32                 }
33                 j--;
34             }
35             
36             if(num ==1){
37                 char[] cArr = new char[bArr.length+1];
38                 for(int l=bArr.length-1;l>=0;l--){
39                     cArr[l+1] = bArr[l];
40                 }
41                 cArr[0] = 1 + '0';
42                 b = String.valueOf(cArr);
43             }else{
44                 b =String.valueOf(bArr);
45             }
46             if(i%10 ==0)
47                 System.out.println();
48             System.out.print(" "+b);
49         }
50         
51     }

②、找大数组中的最大的N位

 1 /*
 2      *    a为源数组，b为要找最大的多少位 
 3      */
 4     static int[] method(int a[], int b){
 5         
 6         int[] nums = new int[b];
 7         
 8         for(int i = 0;i<b;i++)
 9             nums[i] = a[i];
10         
11         Arrays.sort(nums);
12         
13         for(int i = b;i<a.length;i++){
14             
15             if(a[i]>nums[b-1]){
16                 int j = b-1;
17                 while(j>0&&a[i]>nums[j]){
18                     nums[j] = nums[j-1];
19                     j--;
20                 }
21                     
22                 nums[j] = a[i];
23                 
24             }
25         }
26         
27         return nums;
28         
29     }