1，问题描述
jack每天同时只能在A和B其中一个城市工作赚钱，假设两个城市间的交通费为m。已知每天在A 和 B 能赚到多少钱，那么jack怎么选择每天工作的城市才能赚到最大利润。
比如 moneyA = {1，2，3，4，3，1}；moneyB = {2，1，3，4，2，1}；m = 1。
2，边界条件：无
3，思路：假设某一天jack在A城市工作，那么接下来怎么选择到最后一个才能赚到最多，在B城市类似。然后形成递归。
4，代码实现
1）简单递归
1 public static void main(String[] args) { 2 int[] cityA = {1,2,4,5,6,3,1}; 3 int[] cityB = {2,3,4,5,1,1,2}; 4 int res = Math.max(mostMoney(cityA, cityB, 1, 0), mostMoney(cityB, cityA, 1, 0)); 5 System.out.println("res: " + res); 6 } 7 8 public static int mostMoney(int[] curCity, int[] nextCity, int m, int idx) { 9 if (idx >= curCity.length) { 10 return 0; 11 } 12 13 int moneyCur = mostMoney(curCity, nextCity, m, idx + 1); 14 int moneyNext = mostMoney(nextCity, curCity, m, idx + 1) - m; 15 return curCity[idx] + Math.max(moneyCur, moneyNext); 16 }
2）其实在递归时在某一个点能获取的最大利润重复计算，所以获得该最大利润后需记录下来cache，下次直接用，这就是动态规划DP。
1 private static int[] recA; 2 private static int[] recB; 3 public static void main(String[] args) { 4 int[] cityA = {1,2,4,5,6,3,1}; 5 int[] cityB = {2,3,4,5,1,1,2}; 6 recA = new int[cityA.length]; 7 recB = new int[cityB.length]; 8 recA[0] = mostMoney(cityA, cityB, true, 1, 0); 9 recB[0] = mostMoney(cityB, cityA, false, 1, 0); 10 int res = Math.max(recA[0], recB[0]); 11 System.out.println("res: " + res); 12 System.out.println("recA: " + Arrays.toString(recA)); 13 System.out.println("recB: " + Arrays.toString(recB)); 14 } 15 16 public static int mostMoney(int[] curCity, int[] nextCity, boolean isCityA, int m, int idx) { 17 if (idx >= curCity.length) { 18 return 0; 19 } 20 if (isCityA && recA[idx] != 0) { 21 return recA[idx]; 22 } else if (recB[idx] != 0) { 23 return recB[idx]; 24 } 25 26 int moneyCur = mostMoney(curCity, nextCity, isCityA, m, idx + 1); 27 int moneyNext = mostMoney(nextCity, curCity, !isCityA, m, idx + 1) - m; 28 int mostCur = curCity[idx] + Math.max(moneyCur, moneyNext); 29 if (isCityA == true) { 30 recA[idx] = mostCur; 31 } else { 32 recB[idx] = mostCur; 33 } 34 35 return mostCur; 36 }
3）记录下来之后，写成非递归形式
1 private static int[] recA; 2 private static int[] recB; 3 public static void main(String[] args) { 4 int[] cityA = {1,2,4,5,6,3,1}; 5 int[] cityB = {2,3,4,5,1,1,2}; 6 int m = 1; 7 recA = new int[cityA.length]; 8 recB = new int[cityB.length]; 9 recA[cityA.length - 1] = cityA[cityA.length - 1]; 10 recB[cityB.length - 1] = cityB[cityB.length - 1]; 11 for (int i = cityA.length - 2; i >= 0; i--) { 12 recA[i] = cityA[i] + Math.max(recA[i + 1], recB[i + 1] - m); 13 recB[i] = cityB[i] + Math.max(recA[i + 1] - m, recB[i + 1]); 14 } 15 int res = Math.max(recA[0], recB[0]); 16 System.out.println("res: " + res); 17 System.out.println("recA: " + Arrays.toString(recA)); 18 System.out.println("recB: " + Arrays.toString(recB)); 19 }
5，时间复杂度：非递归形式是O（n）
6，api：无