First Missing Positive

Problem:
Given an unsorted integer array, find the first missing positive integer.
For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.
Your algorithm should run in O(n) time and uses constant space.

Solution:
class Solution {
public:
    int firstMissingPositive(int A[], int n) {
        for (int cur = 0; cur < n; cur++) {
            if (A[cur] <= 0) {
                A[cur] = n + 1;
            }
        }

        for (int i = 0; i < n; i++)
            if (abs(A[i]) <= n && A[abs(A[i]) - 1] > 0)
                A[abs(A[i]) - 1] *= -1;

        for (int i = 0; i < n; i++)
            if (A[i] > 0) {
                return i + 1;
            }

        return n + 1;
    }
};

Maximum Subarray

Problem:

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

Solution:

class Solution {
public:
    int maxSubArray(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (n < 1)
            return 0;
       
        int max_sum = A[0];
        int sum_so_far = 0;
        for (int i = 0; i < n; i++) {
            sum_so_far += A[i];
            max_sum = max(max_sum, sum_so_far);
            if (sum_so_far < 0)
                sum_so_far = 0;
        }
        return max_sum;
    }
};

Path Sum

Problem:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

Solution:


/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL)
            return false;
        if (root != NULL && root->left == NULL &&
            root->right == NULL && sum == root->val)
            return true;
        else {
            return hasPathSum(root->left, sum - root->val) ||
                hasPathSum(root->right, sum - root->val);
        }
    }
};

====== Second Try ========
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL)
            return false;
        else if (root -> val == sum && root -> left == NULL && root -> right == NULL)
            return true;
        else if (hasPathSum(root -> left, sum - root -> val) ||
                 hasPathSum(root -> right, sum - root -> val))
            return true;
        else
            return false;
    }
};

==== Clean one ====

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL)
            return false;
        else if (root -> val == sum && root -> left == NULL && root -> right == NULL)
            return true;
        else
            return hasPathSum(root -> left, sum - root -> val) ||
                 hasPathSum(root -> right, sum - root -> val);
    }
};