149. Best Time to Buy and Sell Stock

public class Solution {    /**     * @param prices: Given an integer array     * @return: Maximum profit     */    public int maxProfit(int[] prices) {        // write your code here        if(prices==null || prices.length==0){            return 0;        }                int min = Integer.MAX_VALUE;        int res = Integer.MIN_VALUE;        for(int i=0;i

 

150. Best Time to Buy and Sell Stock II

class Solution {    public int maxProfit(int[] prices) {        if (prices == null || prices.length == 0) {            return 0;        }        int res = 0;        for (int i = 0; i < prices.length - 1; i++) {            if (prices[i + 1] - prices[i] > 0) {                res += prices[i + 1] - prices[i];            }        }        return res;    }}

 

151. Best Time to Buy and Sell Stock III

public class Solution {    /**     * @param prices: Given an integer array     * @return: Maximum profit     */    public int maxProfit(int[] prices) {        // write your code here        if(prices==null||prices.length==0){            return 0;        }        int n= prices.length;        int[][] f = new int[n+1][5+1];                f[0][0]=0;                for(int i=1;i<=n;i++){            //手中无股票            for(int j=1;j<=5;j+=2){                //情况1:昨日卖出                f[i][j] = Math.max(f[i][j],f[i-1][j]);                //情况2:昨日持有+今日卖出收益                if(j>1 && i>1){                    f[i][j] = Math.max(f[i][j],f[i-1][j-1]+prices[i-1]-prices[i-2]);                }            }                        //手中有股票            for (int j = 2; j <= 2 * 2; j += 2) {                // two option: sell or don't sell                if (i > 1) {                    //情况1：昨日持有+今日收益                    f[i][j] = Math.max(f[i][j], f[i - 1][j] + (prices[i - 1] - prices[i - 2]));                }                if (j > 1) {                    //情况2:昨日卖出                    f[i][j] = Math.max(f[i][j], f[i - 1][j - 1]);                }            }        }                int res = Integer.MIN_VALUE;        for (int j = 1; j <= 5; j += 2) {            res = Math.max(res, f[n][j]);        }                return res;    }}

 

393. Best Time to Buy and Sell Stock IV

public class Solution {    /**     * @param K: An integer     * @param prices: An integer array     * @return: Maximum profit     */    public int maxProfit(int K, int[] prices) {        // write your code here        int n = prices.length;        if(n==0){            return 0;        }                if(K>n/2){            int tmp =0;            for(int i=0;i
      
       =2 && j>1){  //注意要加入j>1的边界，j=1时，前一状态是不可能持有股票                    //昨日持有，今日卖出                    f[i][j] = Math.max(f[i][j],f[i-1][j-1]+prices[i-1]-prices[i-2]);                }            }                        //持有股票            for(int j =2;j<=2*K;j+=2){                //昨日已持有，需累加今日收益                if(i>=2)                f[i][j] = Math.max(f[i][j],f[i-1][j]+prices[i-1]-prices[i-2]);                //昨日不持有，今日刚买入                f[i][j] = Math.max(f[i][j],f[i-1][j-1]);            }        }                int res = 0;        for(int j =1;j<=2*K+1;j+=2){            res = Math.max(res,f[n][j]);        }                return res;    }}