HouseRobber(leetcode198)

ou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

/**
 * 這個系統給我返回了Time Limit Exceeded 怎麼能夠優化一下呢 是否是由於複製數組浪費空間 這裏用了遞歸
 * 我想的是在偷第n家的，他不是能夠知道前面怎麼偷的嗎 那邊就有點像斐波那契數列
 */
public static int rob(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    if (nums.length == 1) {
        return nums[0];
    }
    if (nums.length == 2) {
        return nums[0] > nums[1] ? nums[0] : nums[1];
    }
    int[] numsnew = new int[nums.length - 1];
    int[] numsnew2 = new int[nums.length - 2];
    System.arraycopy(nums, 0, numsnew, 0, nums.length - 1);
    System.arraycopy(nums, 0, numsnew2, 0, nums.length - 2);
    int a = rob(numsnew);
    int b = rob(numsnew2) + nums[nums.length - 1];
    return a > b ? a : b;
}

/*
  那麼就迭代一下吧 慢慢向前推導
 */
public static int rob2(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    int[] numsnew = new int[nums.length];
    if(nums.length > 0) {
        numsnew[0] = nums[0];
    }
    if(nums.length > 1) {
        numsnew[1] = nums[0] > nums[1] ? nums[0] : nums[1];
    }
    int length = nums.length;
    int i =2;
    while(i<length){
        numsnew[i] = (numsnew[i-1] > numsnew[i-2]+nums[i])?numsnew[i-1]:numsnew[i-2]+nums[i];
        i++;
    }
    return numsnew[length-1];
}