LintCode(M)打劫房屋2——动态规划

思路：
最后一次的状态方程为：
i=nums.size()-1;//i为最后一家
a[i]=max(houseRobber(A)+nums[i], houseRobber(B));
说明：1. houseRobber(A)+nums[i]为抢最后一家的情况，
A为{nums[1]….nums[i-2]}//少了第一家和最后两家。
2. houseRobber(A)+nums[i]为不抢最后一家的情况，
B为{nums[0]…nums[i-1]}//少了开头第一家。
3.houseRobber是打劫房屋1的代码。
代码：

class Solution {
public:
    /**
     * @param nums: An array of non-negative integers.
     * return: The maximum amount of money you can rob tonight
     */
  long long houseRobber(vector<int> A) {
    // write your code here
    // int n = A.size();
    vector<long long> a(A.size(), 0);
    long long temp1, temp2;
    if (A.size() == 1) return A[0];
    if (A.size() == 2) return max(A[0], A[1]);
    a[0] = 0; a[1] = A[1];

    for (int i = 2; i<A.size(); ++i){
        a[i] = max(a[i - 2] + A[i], a[i - 1]);
    }
    temp1 = a[A.size() - 1];
    a[0] = A[0]; a[1] = 0;
    for (int i = 2; i<A.size(); ++i){
        a[i] = max(a[i - 2] + A[i], a[i - 1]);
    }
    temp2 = a[A.size() - 1];
    return max(temp1, temp2);

}

int houseRobber2(vector<int>& nums) {
    //for (auto c : nums)cout << c << endl;
    if (nums.size() == 1)return nums[0];
    if (nums.size() == 2)return 0;
    if (nums.size() == 3)return nums[1];
    int temp1, temp2;
    vector<int>::iterator it = nums.end(), it2,it3; 
    int b = *(--it);
    it2=nums.erase(it);
    temp1 = houseRobber(nums); 
    //for (auto c : nums)cout << c << endl;
    //不打劫
    //cout << temp1 << endl;    
     nums.erase(--it2);
    it = nums.begin();
    nums.erase(it);
    //for (auto c : nums)cout << c << endl;
    temp2 = houseRobber(nums);
    //打劫
    //cout << temp2 << endl;
    temp2 += b;
    int ret = max(temp1, temp2);

//  for (auto c : nums)cout << c<<endl;
    return ret;
}

};

这个错误结果让我发现我通过lintcode的打劫房屋1的代码有误，就打劫房屋1的代码而言，当我的数据为[4,1,2,7,5,3]时，我的输出答案是11，但是正确答案应该是14，打劫房屋1的思路是正确的，但是我在打劫房屋1里有一个误解：两栋相隔房屋之间必有一家被打劫，这是错误的。
修改打劫房屋1的代码后，全部代码成功：

class Solution {
public:
    /**
     * @param nums: An array of non-negative integers.
     * return: The maximum amount of money you can rob tonight
     */
 long long houseRobber(vector<int> A) {
    // write your code here
    // int n = A.size();
    vector<long long> a(A.size(), 0);
    long long temp1, temp2;
    if (A.size() == 1) return A[0];
    if (A.size() == 2) return max(A[0], A[1]);
    /*a[0] = 0; a[1] = A[1];

    for (int i = 2; i<A.size(); ++i){
        a[i] = max(a[i - 2] + A[i], a[i - 1]);
    }
    temp1 = a[A.size() - 1];
    a[0] = A[0]; a[1] = 0;*/
    a[0] = A[0]; a[1] = max(A[0], A[1]);
    for (int i = 2; i<A.size(); ++i){
        a[i] = max(a[i - 2] + A[i], a[i - 1]);
    }
    temp2 = a[A.size() - 1];
    return temp2;
    //return max(temp1, temp2);

}

int houseRobber2(vector<int>& nums) {
    //for (auto c : nums)cout << c << endl;
    if (nums.size() == 1)return nums[0];
    if (nums.size() == 2)return 0;
    if (nums.size() == 3)return nums[1];
    int temp1, temp2;
    vector<int>::iterator it = nums.end(), it2,it3; 
    int b = *(--it);
    it2=nums.erase(it);
    temp1 = houseRobber(nums); 
    //for (auto c : nums)cout << c << endl;
    //不打劫
    //cout << temp1 << endl;    
     nums.erase(--it2);
    it = nums.begin();
    nums.erase(it);
    //for (auto c : nums)cout << c << endl;
    temp2 = houseRobber(nums);
    //打劫
    //cout << temp2 << endl;
    temp2 += b;
    int ret = max(temp1, temp2);

//  for (auto c : nums)cout << c<<endl;
    return ret;
}

};