【LEETCODE】69、动态规划，easy,medium级别，题目：198、139、221

package y2019.Algorithm.dynamicprogramming.easy;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.dynamicprogramming.easy
 * @ClassName: Rob
 * @Author: xiaof
 * @Description: 198. House Robber
 * 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.
 *
 * Example 1:
 *
 * Input: [1,2,3,1]
 * Output: 4
 * Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
 *              Total amount you can rob = 1 + 3 = 4.
 * Example 2:
 *
 * Input: [2,7,9,3,1]
 * Output: 12
 * Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
 *              Total amount you can rob = 2 + 9 + 1 = 12.
 * @Date: 2019/8/16 8:45
 * @Version: 1.0
 */
public class Rob {

    public int solution(int[] nums) {
        if (nums == null || nums.length <= 0) {
            return 0;
        }
        //这题主要就是要发现规律，那就是选择盗窃的时候，当前室是否要抢劫进入
        //那么区别就是rob(i) = max{rob(i-2) + curhouse, rob(i-1)}
        int[] dp = new int[nums.length + 1];
        dp[1] = nums[0];
        for (int i = 2; i < dp.length; ++i) {
            dp[i] = Math.max(dp[i - 2] + nums[i - 1], dp[i - 1]);
        }

        return dp[nums.length];
    }
}

 

 

package y2019.Algorithm.dynamicprogramming.medium;

import java.util.List;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.dynamicprogramming.medium
 * @ClassName: WordBreak
 * @Author: xiaof
 * @Description: 139. Word Break
 * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words,
 * determine if s can be segmented into a space-separated sequence of one or more dictionary words.
 *
 * Note:
 *
 * The same word in the dictionary may be reused multiple times in the segmentation.
 * You may assume the dictionary does not contain duplicate words.
 * Example 1:
 *
 * Input: s = "leetcode", wordDict = ["leet", "code"]
 * Output: true
 * Explanation: Return true because "leetcode" can be segmented as "leet code".
 * Example 2:
 *
 * Input: s = "applepenapple", wordDict = ["apple", "pen"]
 * Output: true
 * Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
 *              Note that you are allowed to reuse a dictionary word.
 * Example 3:
 *
 * Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
 * Output: false
 * @Date: 2019/8/16 8:46
 * @Version: 1.0
 */
public class WordBreak {

    public boolean solution(String s, List<String> wordDict) {
        //我们把s当做一个地址的字符数组，每次获取到一个新的字符的时候
        //判断是否可以和字典匹配成功，如果成功那么当前位置的值就是true（任意一个字符串）
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true; //如果字符串长度为0，那么默认为true

        for (int i = 1; i < dp.length; ++i) {
            //遍历所有的字符,进行比较
            for (String curs : wordDict) {
                //比较当前字符位置
                if (i >= curs.length()) {
                    //前面的字符也要比较确认没问题才能继续比较
                    if (dp[i - curs.length()]) {
                        //获取相同长度的字符进行比较，然后把前面剩下的字符再进行比较
                        String compare = s.substring(i - curs.length(), i);
                        if (compare.equals(curs)) {
                            dp[i] = true;
                            break;
                        }
                    }
                }
            }

        }

        return dp[s.length()];
    }
}

 

package y2019.Algorithm.dynamicprogramming.medium;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.dynamicprogramming.medium
 * @ClassName: MaximalSquare
 * @Author: xiaof
 * @Description: 221. Maximal Square
 * Given a 2D binary matrix filled with 0's and 1's, find the largest square(正方形) containing only 1's and return its area.
 *
 * Example:
 *
 * Input:
 *
 * 1 0 1 0 0
 * 1 0 1 1 1
 * 1 1 1 1 1
 * 1 0 0 1 0
 *
 * Output: 4
 * @Date: 2019/8/16 8:46
 * @Version: 1.0
 */
public class MaximalSquare {

    public int solution(char[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return 0;
        }
        //这题可以转换为求边长，只有一个点的上面，左边，和左上角都是1的时候，才进行长度加1，如果有一个反向的值不为1，那么就无法进行加一
        //这个二维数组用来求边长
        int[][] dp = new int[matrix.length + 1][matrix[0].length + 1];
        int maxlen = 0;

        for (int i = 1; i < dp.length; ++i) {
            for (int j = 1; j < dp[i].length; ++j) {
                //首先判断当前位置是否为空，如果是那么就判断边长能否添加
                if (matrix[i - 1][j - 1] == '1') {
                    dp[i][j] = Math.min(dp[i-1][j], Math.min(dp[i][j-1], dp[i-1][j-1])) + 1;
                    maxlen = Math.max(maxlen, dp[i][j]);
                }
            }
        }

        return maxlen * maxlen;
    }
}