LeetCode 122, 123, 188. Best Time to Buy and Sell Stock II+III+IV

Best Time to Buy and Sell Stock II

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res=0;
        for (int i=1;i<prices.size();++i){
            if (prices[i]>prices[i-1]){
                res += prices[i]-prices[i-1];
            }
        }
        return res;
    }
};

Best Time to Buy and Sell Stock III

比较难想到，分别记录四个状态: 1buy，1buy1sell，2buy1sell，2buy2sell。

构建四个dp数组，例如，1buy1sell[i] = max( 1buy1sell[i-1], 1buy[i-1]+prices[i] )

由于dp只依赖于上一个状态，可以压缩空间。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n=prices.size();
        int dp[2][4]; // 0 - 1 buy, 1 - 1 buy 1 sell, 2 - 2 buy 1 sell, 3 - 2 buy 2 sell
        for (int i=0;i<4;++i)
            dp[0][i] = INT_MIN/2;

        for (int i=0;i<n;++i){
            dp[1][0] = max(dp[0][0],-prices[i]);
            dp[1][1] = max(dp[0][1],dp[0][0]+prices[i]);
            dp[1][2] = max(dp[0][2],dp[0][1]-prices[i]);
            dp[1][3] = max(dp[0][3],dp[0][2]+prices[i]);
            swap(dp[0],dp[1]);
        }
        return max(0, max(dp[0][1],dp[0][3]));
    }
};

Best Time to Buy and Sell Stock IV

上一题的拓展，思路一样。corner case 比较多，需要特别注意。当 k > n/2 的时候，说明可以不限次数，本题就变成了 Best Time to Buy and Sell Stock II，使用 II 的方法，可以大大减少dp所需的时间和空间。

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (k==0) return 0;
        int n=prices.size();

        if (n<2) return 0;
        if (k>n/2){ // Best Time to Buy and Sell Stock II
            int ans = 0;
            for (int i=1; i<n; ++i)
                ans += max(prices[i] - prices[i-1],0);
            return ans;
        }

        // 0 - 1 buy, 1 - 1 buy 1 sell, 2 - 2 buy 1 sell, 3 - 2 buy 2 sell ...
        vector<vector<int>> dp(2,vector<int>(2*k));
        for (int j=0;j<2*k;++j)
            dp[0][j] = INT_MIN/2;

        for (int i=0;i<n;++i){
            dp[1][0] = max(dp[0][0],-prices[i]);
            for (int j=1;j<2*k;++j){
                dp[1][j] = max( dp[0][j], dp[0][j-1]+(j%2==1?1:-1)*prices[i] );
            }
            swap(dp[0],dp[1]);
        }
        int res=0;
        for (int j=1;j<2*k;j+=2)
            res = max(res,dp[0][j]);
        return res;
    }
};

原文地址：https://www.cnblogs.com/hankunyan/p/11258505.html