各种算法题MARK

①、斐波那契数列  

static void method(){
        
        String a = "1",b="1";
        char[] aArr;
        char[] bArr;
        int num = 0;
        int i,j;
        int temp = 0;
        int k;
        System.out.print(" "+a+" "+b);
        for(i =2 ; i<=500;i++){
            aArr = a.toCharArray();
            bArr = b.toCharArray();
            a = b;
            k = bArr.length - aArr.length;
            for(j = aArr.length-1;j>=0;j--){
                temp = (bArr[j+k]-'0')+(aArr[j]-'0') + num;
                num = 0;
                if(temp >=10){
                    temp -=10;
                    num = 1;
                }
                bArr[j+k] = (char) (temp + '0');
            }
            while(j+k>=0 &&num==1){
                temp = (bArr[j+k] - '0') + num;
                if(temp > 10){
                    bArr[j+k] = (char) (bArr[j+k] + (temp - 10));
                    num = 1;
                }else{
                    num = 0;
                }
                j--;
            }
            
            if(num ==1){
                char[] cArr = new char[bArr.length+1];
                for(int l=bArr.length-1;l>=0;l--){
                    cArr[l+1] = bArr[l];
                }
                cArr[0] = 1 + '0';
                b = String.valueOf(cArr);
            }else{
                b =String.valueOf(bArr);
            }
            if(i%10 ==0)
                System.out.println();
            System.out.print(" "+b);
        }
        
    }

②、找大数组中的最大的N位

/*
     *    a为源数组，b为要找最大的多少位 
     */
    static int[] method(int a[], int b){
        
        int[] nums = new int[b];
        
        for(int i = 0;i<b;i++)
            nums[i] = a[i];
        
        Arrays.sort(nums);
        
        for(int i = b;i<a.length;i++){
            
            if(a[i]>nums[b-1]){
                int j = b-1;
                while(j>0&&a[i]>nums[j]){
                    nums[j] = nums[j-1];
                    j--;
                }
                    
                nums[j] = a[i];
                
            }
        }
        
        return nums;
        
    }