Lintcode 366 Fibonacci

/* 1st method would lead to time limit */
/* The time complexity is exponential sicne t (n) = t (n-1) + t (n-2) */

Class Solution {    /**     * @param n:an integer     * @return an integer f (n) */public    int Fibonacci (int n) {        //Write your code here        if (n = = 1 | | n = = 2) {            return (n-1);        }                int sum = (n-1) + (n-2);                Return Fibonacci (n-1) + Fibonacci (n-2);            }}

/* 2nd method would need O (n) space, using DP */
/* T and S are both O (n) */

1  Public intFibonacciintN) {2         //declare an array to store the result3         //It have to is n+2 to avoid out_of_bound4         int[] f =New int[N+2]; 5F[1] = 0;//When input is 1 = zero6F[2] = 1;//When input is 2 = 17         8         inti = 3;9          while(I <= N) {//it has been incremental instead of DecrementalTenF[i] = f[i-1] + f[i-2]; Onei++; A         } -          -         returnF[n]; the}

/* 3rd method would only need O (1) space */
/* We can optimize the space used in method 2 by storing the previous and numbers only */
/* Because that's all we need to get the next Fibannaci number in series. */

1  Public intFibonacciintN) {2     if(N < 3)returnN-1;3      4     intFirst = 0;5     intSecond = 1;6     intThird = 1;7     8     inti = 3;9      while(I <=N) {TenThird = first +second; OneFirst =second; ASecond =third; -i++; -     } the     returnthird; -}

Lintcode 366 Fibonacci