@[toc]

一、I. 斐波那契数列

    public int fib(int n) {
        int a = 0, b = 1, sum;
        for(int i = 0; i < n; i++){
            sum = (a + b) % 1000000007;
            a = b;
            b = sum;
        }
        return a;
    }

二、II. 青蛙跳台阶问题

    public int numWays(int n) {
        if (n == 0 || n == 1) {
            return 1;
        }
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i < dp.length; i++) {
            dp[i] = (dp[i - 1] + dp[i - 2]) % 1000000007;
        }
        return dp[n];
    }

三、正则表达式匹配

    public boolean isMatch(String s, String p) {
        int m = s.length() + 1, n = p.length() + 1;
        boolean[][] dp = new boolean[m][n];
        dp[0][0] = true;
        for(int j = 2; j < n; j += 2) {
            dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*';
        }
        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
                dp[i][j] = p.charAt(j - 1) == '*' ?
                        dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') :
                        dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1));
            }
        }
        return dp[m - 1][n - 1];
    }

四、连续子数组的最大和

    public int maxSubArray(int[] nums) {
        if (nums.length == 1) {
            return nums[0];
        } else {
            int[] dp = new int[nums.length];
            dp[0] = nums[0];
            int max = dp[0];
            for (int i = 1; i < dp.length; i++) {
                dp[i] = dp[i - 1] > 0 ? dp[i - 1] + nums[i] : nums[i];
                max = Math.max(dp[i],max);
            }
            return max;
        }
    }

五、把数字翻译成字符串

    public int translateNum(int num) {
        if (num < 10) {
            return 1;
        }
        String str = String.valueOf(num);
        int[] dp = new int[str.length()];
        dp[0] = 1;
        for (int i = 1; i < dp.length; i++) {
            int n = Integer.parseInt(str.substring(i - 1, i + 1));
            if (n <= 25 && str.charAt(i - 1) != '0') {
                if (i > 1) {
                    dp[i] = dp[i - 1] + dp[i - 2];
                } else {
                    dp[i] = dp[i - 1] + 1;
                }
            } else {
                dp[i] = dp[i - 1];
            }
        }
        return dp[str.length() - 1];
    }

六、礼物的最大价值

    public int maxValue(int[][] grid) {
        int[][] dp = new int[grid.length][grid[0].length];
        dp[0][0] = grid[0][0];
        // 初始化第一行
        for (int i = 1; i < dp[0].length; i++) {
            dp[0][i] = grid[0][i] + dp[0][i - 1];
        }
        // 初始化第一列
        for (int i = 1; i < dp.length; i++) {
            dp[i][0] = grid[i][0] + dp[i - 1][0];
        }
        // 开始dp
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                dp[i][j] = grid[i][j] + Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }

七、最长不含重复字符的子字符串

    public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<>();
        int res = 0, tmp = 0, len = s.length();
        for(int j = 0; j < len; j++) {
            // 获取索引 i
            int i = map.getOrDefault(s.charAt(j), -1);
            // 更新哈希表
            map.put(s.charAt(j), j);
            // dp[j - 1] -> dp[j]
            tmp = tmp < j - i ? tmp + 1 : j - i;
            // max(dp[j - 1], dp[j])
            res = Math.max(res, tmp);
        }
        return res;
    }

八、丑数

    public int nthUglyNumber(int n) {
        int a = 0, b = 0, c = 0;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i = 1; i < n; i++) {
            int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;
            dp[i] = Math.min(Math.min(n2, n3), n5);
            if(dp[i] == n2) {
                a++;
            }
            if(dp[i] == n3) {
                b++;
            }
            if(dp[i] == n5) {
                c++;
            }
        }
        return dp[n - 1];
    }

九、n 个骰子的点数

    public double[] dicesProbability(int n) {
        double[] dp = new double[6];
        Arrays.fill(dp, 1.0 / 6.0);
        for (int i = 2; i <= n; i++) {
            double[] tmp = new double[5 * i + 1];
            for (int j = 0; j < dp.length; j++) {
                for (int k = 0; k < 6; k++) {
                    tmp[j + k] += dp[j] / 6.0;
                }
            }
            dp = tmp;
        }
        return dp;
    }

十、股票的最大利润

    public int maxProfit(int[] prices) {
        int cost = Integer.MAX_VALUE, profit = 0;
        for (int price : prices) {
            cost = Math.min(cost, price);
            profit = Math.max(profit, price - cost);
        }
        return profit;
    }