LeetCode_198、213、222三题

LeetCode_198. House Robber

原题：
You 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.

题意大致是，以一个盗贼的身份，偷取一个小镇上的房子里的钱，但不能偷相邻房子，否则会被抓到。给定一个序列，每个元素代表了各个房子的钱数量，求偷取最大的金额。

我们很容易知道，由于不能偷相邻房子的钱，所以可以对序列里相邻的两个数字两两进行操作，比如序号1,2,3,4，假如当前序列是3号，则我们可以加上之前的1号前已经偷到的最多的钱，当前序列是4号，则我们可以加上之前的2号前已经偷到的最多的钱，因此我们只需要两个变量用于存储当前序号前2位偷到的最大金额。

具体实现如下：

int rob(vector<int>& nums) {        int a = 0;        int b = 0;        for(int i = 0;i < nums.size();i++){            if(i % 2 == 0)a = max(a+nums[i],b);            else b = max(b+nums[i],a);        }        return max(a,b);    }

LeetCode_213.House Robber II

原题：
After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

这一题是上一题的延伸，只是把表的最后一个元素与第一个元素也当成相邻的了（即表成为了一个环状）。所以，当取最后一个元素时，不能取第一个元素。于是我们可以分别对表中除了最后一个元素的序列以第一题的方式处理一次，再对除了第一个元素的序列以第一题的方式再处理一次，然后求两种情况下的最大值就是结果。具体实现如下：

int rob(vector<int>& nums) {        int a = 0;        int b = 0;        if(nums.size() == 0)return 0;        if(nums.size() == 1)return nums[0];        for(int i = 0;i < nums.size() - 1;i ++){            if(i % 2)a = max(a+nums[i],b);            else b = max(b+nums[i],a);        }        int sum1 = max(a,b);        a = 0;        b = 0;        for(int i = 1;i < nums.size();i ++){            if(i % 2)a = max(a+nums[i],b);            else b = max(b+nums[i],a);        }        int sum2 = max(a,b);        return max(sum1,sum2);    }

LeetCode_55. Jump Game
原题：
Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4] return true.
A = [3,2,1,0,4], return false.

题意很简单，就是判断我们能不能到达最后一个元素。
所以首先确定当序列只有一个元素时，本身就是最后一个元素，所以可以到达。
其次，当第一个元素是0且序列大小大于1时，必定不可以到达。
最后考虑一般情况，一般情况下，从后往前，只需要找到为0的元素，若在它之前有元素可以跳到它的后面，则可以到达到0的后面，否则一定不可以跳过这个0，也就无法到达最后一个元素。

具体代码如下：

bool canJump(vector<int>& nums) {        if(nums[0] == 0 && nums.size() > 1)return false;        for(int i = nums.size() - 1;i >= 0;i --){            if(i == nums.size() - 1 && nums[i] == 0){                for(int j = 0; j < i; j++){                    if(nums[j] + j >= i){                        i = j;                        break;                    }                    if(j == i - 1)return false;                }            }            if(nums[i] == 0){                for(int j = 0; j < i; j++){                    if(nums[j] + j > i){                        i = j;                        break;                    }                    if(j == i - 1)return false;                }            }        }        return true;    }