Monday, January 11, 2016

Lintcode Fibonacci - Bogart2015 - 博客园


Lintcode Fibonacci - Bogart2015 - 博客园
Find the Nth number in Fibonacci sequence.
A Fibonacci sequence is defined as follow:

  • The first two numbers are 0 and 1.

  • The i th number is the sum of i-1 th number and i-2 th number.
    The first ten numbers in Fibonacci sequence is:
    0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ...
    Example
    Given 1, return 0
    Given 2, return 1
    Given 10, return 34
    Note
    The Nth fibonacci number won't exceed the max value of signed 32-bit integer in the test cases.
    In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
        Fn = Fn-1 + Fn-2
    with seed values
       F1 = 0 and F2 = 1
    int fibonacci(int n) { 
          if (n == 1)
          return 0;
          else if (n == 2)
          return 1;
          return fib(n-1) + fib(n-2);
    }
    Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential. So this is a bad implementation for nth Fibonacci number.
    Extra Space: O(n) if we consider the function call stack size, otherwise O(1).
    Method 2 ( Use Dynamic Programming )
    int fibonacci(int n)
    {
    /* Declare an array to store Fibonacci numbers. */
      int f[n+1];
      int i; 
      /* 0th and 1st number of the series are 0 and 1*/
      f[1] = 0;
      f[2] = 1;
      for (i = 2; i <= n; i++)
      {
          /* Add the previous 2 numbers in the series and store it */
          f[i] = f[i-1] + f[i-2];
      } 
      return f[n];
    }
    Method 3 ( Space Otimized Method 2)
    We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibannaci number in series.
    int fibonacci(int n)
    {
      int a = 0, b = 1, c, i;
      if( n == 1) return a; 
      if( n == 2) return b; 
      for (i = 2; i <= n; i++)
      {
         c = a + b;
         a = b;
         b = c;
      }
      return b;
    }
    http://cherylintcode.blogspot.com/2015/06/fibonacci.html
        public int fibonacci(int n) {
            // write your code here
            if(n <= 2) return n - 1;
            int pre = 0, cur = 1;
            for(int i = 3; i <= n; i++){
                int temp = cur;
                if(Integer.MAX_VALUE - pre < cur) return Integer.MAX_VALUE;
                cur = pre +cur;
                pre = temp;
            }
            return cur;
        }
    http://blog.csdn.net/willshine19/article/details/46907199
        HashMap<Integer, Integer> map;
        public int fibonacci(int n) {
            // 2015-07-16 
            // 用哈希表提高时间效率,算过的值就保存下来,不用再算
            // 注意递归要向“已知”的方向
            if (map == null) {
                map = new HashMap<>();
            }
            if (n <= 0) {
                return -1;
            } else if (n == 1) {
                return 0;
            } else if (n == 2) {
                return 1;
            } else {
                if (!map.containsKey(n - 1)) {
                    map.put(n - 1, fibonacci(n - 1));
                }
                if (!map.containsKey(n - 2)) {
                    map.put(n - 2, fibonacci(n - 2));
                }
                return map.get(n - 1) + map.get(n - 2);
            }        
        }
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