LeetCode 121-125

121. 买卖股票的最佳时机

详细题解见动态规划

class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 0) {
            return 0;
        }
        int[][] dp = new int[prices.length][2];
        dp[0][0] = 0;
        dp[0][1] = -prices[0];
        for (int i = 1; i < prices.length; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
            dp[i][1] = Math.max(dp[i - 1][1], 0 - prices[i]);
        }
        return dp[prices.length - 1][0];
    }
}

122. 买卖股票的最佳时机 II

class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if (len == 0) {
            return 0;
        }
        int[][] dp = new int[len][2];
        dp[0][0] = 0;
        dp[0][1] = -prices[0];
        for (int i = 1; i < len; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
            dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
        }
        return dp[len - 1][0];
    }
}

123. 买卖股票的最佳时机 III

class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if (len == 0) {
            return 0;
        }
        int[][][] dp = new int[len][3][2];
        for (int i = 1; i <= 2; i++) {
            dp[0][i][0] = 0;
            dp[0][i][1] = -prices[0];
        }
        for (int i = 1; i < len; i++) {
            for (int j = 1; j <= 2; j++) {
                dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
                dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
            }
        }
        return dp[len - 1][2][0];
    }
}

124. 二叉树中的最大路径和

class Solution {
    int res = Integer.MIN_VALUE;

    public int maxPathSum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        dfs(root);
        return res;
    }

    public int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = Math.max(0, dfs(root.left));//左孩子贡献
        int right = Math.max(0, dfs(root.right));//右孩子贡献
        res = Math.max(res, root.val + left + right);
        return root.val + Math.max(left, right);
    }
}

125. 验证回文串

class Solution {
    public boolean judge(char c) {
        if (Character.isDigit(c) || Character.isLetter(c)) {
            return true;
        } else {
            return false;
        }
    }

    public boolean isPalindrome(String s) {
        if ("".equals(s)) {
            return true;
        }
        int i = 0, j = s.length() - 1;
        while (i < j) {
            while (i < s.length() && judge(s.charAt(i)) == false) {
                i++;
            }
            while (j >= 0 && judge(s.charAt(j)) == false) {
                j--;
            }
            if (i >= j) {
                break;
            }
            if (Character.toLowerCase(s.charAt(i++)) != Character.toLowerCase(s.charAt(j--))) {
                return false;
            }
        }
        return true;
    }
}