标签：nbsp ace for lint tmp ber return lin tco
// Ref: https://segmentfault.com/a/1190000003811581
// Ref: http://www.cnblogs.com/grandyang/p/4383632.html
/*
如果选择了抢劫上一个屋子，那么就不能抢劫当前的屋子，所以最大收益就是抢劫上一个屋子的收益
如果选择抢劫当前屋子，就不能抢劫上一个屋子，所以最大收益是到上一个屋子的上一个屋子为止的最大收益，加上当前屋子里有的钱
*/
1 public static long houseRobber(int[] A) { 2 int len = A.length; 3 if (len <= 1) { 4 return len == 0 ? 0: A[0]; 5 } 6 7 long previous = A[0]; // a is the previous max 8 long current = Math.max(A[0], A[1]); // b is the current max 9 for(int i = 2; i < len; i++){ 10 long tmp = current; 11 // previous + A[i] => pre-pre + current 12 // b/c it cannot rob adjacent houses => need to compare 13 current = Math.max(previous + A[i], current); 14 previous = tmp; 15 } 16 return current; 17 }
标签：nbsp ace for lint tmp ber return lin tco
原文地址：http://www.cnblogs.com/wenchan/p/6209645.html
