斐波那契问题（Java实现）

具体思路参考代码面试指南P182

package go.jacob.day1201;/** * 斐波那契数列 *  * @author Administrator 记住两个方法：1.O(n)时间复杂度用循环; 2.O(logn)用矩阵相乘 切记不要用递归 */public class Demo2 {/* * 方法一：循环 时间复杂度O(n) */public int Fibonacci_1(int n) {if (n < 1)return 0;if (n == 1 || n == 2)return 1;int res = 1;int pre = 1;int tmp = 0;for (int i = 3; i <= n; i++) {tmp = res;res = res + pre;pre = tmp;}return res;}/* * 结论：F(n)=F(n-1)+F(n-2),是一个二阶递推数列， * 一定可以用矩阵乘法的形式表示  * 这道题的递推矩阵为[1,1;1,0] */public int Fibonacci_2(int n) {if (n < 1)return 0;if (n == 1 || n == 2)return 1;int[][] base = { { 1, 1 }, { 1, 0 } };int[][] res = maxtrixPower(base, n - 2);return res[0][0] + res[0][1];}/* * 求矩阵m的p次方 */private int[][] maxtrixPower(int[][] m, int p) {int[][] res = new int[m.length][m.length];for (int i = 0; i < m.length; i++) {res[i][i] = 1;}int[][] tmp = m;for (; p != 0; p >>= 1) {if ((p & 1) != 0) {res = multiMatrix(res, tmp);}tmp = multiMatrix(tmp, tmp);}return res;}/* * 求两个矩阵相乘 */public int[][] multiMatrix(int[][] m1, int[][] m2) {int[][] res = new int[m1.length][m2[0].length];for (int i = 0; i < m1.length; i++) {for (int j = 0; j < m2[0].length; j++) {for (int k = 0; k < m1[0].length; k++) {res[i][j] += m1[i][k] * m2[k][j];}}}return res;}}