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 ...
Notice
The Nth fibonacci number won't exceed the max value of signed 32-bit integer in the test cases.
Example
Given 1, return 0
Given 2, return 1
Given 10, return 34
解法一：
1 class Solution { 2 /** 3 * @param n: an integer 4 * @return an integer f(n) 5 */ 6 public int fibonacci(int n) { 7 if (n == 1 || n == 2) { 8 return (n-1); 9 } 10 11 return fibonacci(n-1) + fibonacci(n-2); 12 13 } 14 }
will lead to time limit
解法二：
1 class Solution { 2 public int fibonacci(int n) { 3 // declare an array to store the result 4 // it has to be n+2 to avoid out_of_bound 5 int[] f = new int[n+2]; 6 f[1] = 0; // when input is 1 => zero 7 f[2] = 1; // when input is 2 => 1 8 9 int i = 3; 10 while (i <= n) { // it has to be incremental instead of decremental 11 f[i] = f[i-1] + f[i-2]; 12 i++; 13 } 14 15 return f[n]; 16 } 17 }
will need O(n) space, using DP
解法三：
1 class Solution { 2 public int fibonacci(int n) { 3 if (n < 3) { 4 return n - 1; 5 } 6 7 int first = 0; 8 int second = 1; 9 int third = 1; 10 11 int i = 3; 12 while (i <= n) { 13 third = first + second; 14 first = second; 15 second = third; 16 i++; 17 } 18 return third; 19 } 20 }
will only need O(1) space，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