LeetCode 2016 392,337,309,201,416

392 Is Subsequence

class Solution {public:    bool isSubsequence(string s, string t)    {        int ls = s.length(), lt= t.length();        if (ls==0) return true;        int ps=0;        bool flag=false;        for(int i=0;i<lt;i++)        {            if (t[i]==s[ps])            {                ps++;            }            if (ps==ls)            {                flag=true;                break;            }        }        return flag;    }};

337 House Robber III

class Solution {public:    int tryRob(TreeNode* root,int& l,int& r)    {        if (!root) return 0;        int ll=0,lr=0,rl=0,rr=0;        l=tryRob(root->left,ll,lr);        r=tryRob(root->right,rl,rr);        return max(root->val+ll+lr+rl+rr,l+r);    }    int rob(TreeNode* root)    {        int l,r;        return tryRob(root,l,r);    }};

309 Best Time to Buy and Sell Stock with Cooldown

1. 看了Disscuss明显题目中给了三种状态buy，sell，cooldown，就应该想到分成这三个情况来进行动态规划，我个脑残！The series of problems are typical dp. The key for dp is to find the variables to represent the states and deduce the transition function.Of course one may come up with a O(1) space solution directly, but I think it is better to be generous when you think and be greedy when you implement.The natural states for this problem is the 3 possible transactions : buy, sell, rest. Here rest means no transaction on that day (aka cooldown).Then the transaction sequences can end with any of these three states.For each of them we make an array, buy[n], sell[n] and rest[n].buy[i] means before day i what is the maxProfit for any sequence end with buy.sell[i] means before day i what is the maxProfit for any sequence end with sell.rest[i] means before day i what is the maxProfit for any sequence end with rest.Then we want to deduce the transition functions for buy sell and rest. By definition we have:buy[i]  = max(rest[i-1]-price, buy[i-1]) sell[i] = max(buy[i-1]+price, sell[i-1])rest[i] = max(sell[i-1], buy[i-1], rest[i-1])Where price is the price of day i. All of these are very straightforward. They simply represents :(1) We have to `rest` before we `buy` and (2) we have to `buy` before we `sell`One tricky point is how do you make sure you sell before you buy, since from the equations it seems that [buy, rest, buy] is entirely possible.Well, the answer lies within the fact that buy[i] <= rest[i] which means rest[i] = max(sell[i-1], rest[i-1]). That made sure [buy, rest, buy] is never occurred.A further observation is that and rest[i] <= sell[i] is also true thereforerest[i] = sell[i-1]Substitute this in to buy[i] we now have 2 functions instead of 3:buy[i] = max(sell[i-2]-price, buy[i-1])sell[i] = max(buy[i-1]+price, sell[i-1])This is better than 3, butwe can do even better2. class Solution {public:    int maxProfit(vector<int>& prices) {        int buy(INT_MIN),sell(0),prev_sell(0),prev_buy;        int len=prices.size();        for(int i=0;i<len;i++)        {            int price=prices[i];            prev_buy=buy;            buy=max(prev_sell - price,buy);            prev_sell = sell;            sell=max(prev_buy+price,sell);        }        return sell;    }};

201 Bitwise AND of Numbers Range

class Solution {public:    int rangeBitwiseAnd(int m, int n) {        if(m == 0){            return 0;        }        int moveFactor = 1;        while(m != n){            m >>= 1;            n >>= 1;            moveFactor <<= 1;        }        return m * moveFactor;            }};

416 Partition Equal Subset Sum

class Solution {public:    bool canPartition(vector<int>& nums)    {        int len=nums.size();        int sum=0,hsum=0;        for(int i=0;i<len;i++)        {            sum+=nums[i];        }        if (sum%2==1) return false;        hsum=sum/2;        int MAX_F=100*100+10;        bool f[MAX_F];        f[0]=true;        for(int i=0;i<len;i++)        {            for(int j=hsum;j>=nums[i];j--)            {                if (j-nums[i]>=0)                {                    f[j]=f[j]||f[j-nums[i]];                }            }        }        return f[hsum];    }};