java實現——8硬幣問題（演算法）

標籤：8硬幣問題   演算法   java   

代碼看著有點亂，有點複雜！單步調試一下其實不複雜！當然前提是要搞清楚演算法的思想！

package yxd.sf.eightcoins;public class EightCoins {public static int compareTo(int[] coins, int[] left, int[] right) {int leftTotal = 0;int rightTotal = 0;int fakeCoin = 0;int coinsSize = 0;if (coins.length % 2 == 0) {coinsSize = coins.length/2;//將現在的硬幣分成兩半System.arraycopy(coins, 0, left, 0, coinsSize);//前一半硬幣放到leftfor (int i = 0; i < coinsSize; i++) {leftTotal = leftTotal + left[i]; //前一半硬幣的總重量}System.arraycopy(coins, coinsSize, right, 0, coinsSize);//後一半硬幣放到rightfor (int j = 0; j < coinsSize; j++) {rightTotal = rightTotal + right[j];//後一半硬幣的總重量}}//比較right和left兩邊的硬幣重量if (leftTotal < rightTotal && coinsSize != 1) {coins = right;int[] right_1 = new int[coinsSize/2];int[] left_1 = new int[coinsSize/2];right = right_1;//動態改變數組大小，減治後硬幣減半left = left_1;System.arraycopy(coins, 0, left, 0, coinsSize/2);System.arraycopy(coins, coinsSize/2, right, 0, coinsSize/2);} else if (leftTotal > rightTotal && coinsSize != 1) {coins = left;int[] right_1 = new int[coinsSize/2];int[] left_1 = new int[coinsSize/2];right = right_1;left = left_1;System.arraycopy(coins, 0, left, 0, coinsSize/2);System.arraycopy(coins, coinsSize/2, right, 0, coinsSize/2);}if (left.length != 1 && right.length != 1 && fakeCoin == 0) {fakeCoin = compareTo(coins, left, right);    //遞迴調用，如果不令“fakeCoin=”最終返回會為0}if (left.length == 1 && right.length == 1 && coinsSize == 2) {if (left[0] < right[0]) {fakeCoin = right[0];} else if (left[0] == right[0]){System.out.println(left[0] + "and" + right[0]);} else {fakeCoin = left[0];}}return fakeCoin;}public static void main(String[] args) {int[] coins = {1, 1, 1, 2, 1, 1, 1, 1};int[] left = new int[4];int[] right = new int[4];int flag = 0;int location = compareTo(coins, left, right);for (int i = 0; i < coins.length; i++) {flag++;if (coins[i] == location) {System.out.println("第" + flag + "枚硬幣為假幣！");}}}}

java實現——8硬幣問題（演算法）