基础数据结构思路&写法记录,便于回顾

重思路非代码。基础的思路搞懂了,变形题目顺着思考基本都能写出来!以下是自己8年技术面试以来总结的算法考点!

基础(要消化的)

        基础查找 二叉树 链表 排序

二分查找

    int binarySearch(vector<int> &nums, int target) {
        // write your code here
        if (nums.empty()) {
            return -1;
        }

        int start = 0;
        int end = nums.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] >= target) { // 有重复的输出第一个
                end = mid;
            } else {
                start = mid;
            }

            /*
                if (nums[mid] <= target) { // 有重复的输出最后一个
                    start = mid;
                } else {
                    end = mid;
                }
            */
        }

        if (nums[start] == target) {
            return start;
        }

        if (nums[end] == target) {
            return end;
        }

        return -1;
    }

链表逆序

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        
        // 注意的是先画图, 代码自然就能写出来
        ListNode* prev = nullptr;
        ListNode* cur = head;    
        while (cur != nullptr) {
            ListNode* tmp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = tmp;
        }

        return prev;
    }
};

二叉树遍历

// 前序
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        if (root == nullptr) {
            return {};
        }

        stack<TreeNode*> s;
        // s.push(root);
        vector<int> res;
        while (!s.empty() || root) {
            if (root) {
                res.push_back(root->val);
                s.push(root);
                root = root->left;
            } else {
                root = s.top();
                s.pop();
                root = root->right;
            }
        }

        return res;
    }
};

// 中序
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        if (root == nullptr) {
            res;
        }

        stack<TreeNode*> s;
        // s.push(root);

        while (!s.empty() || root) {
            if (root) {
                s.push(root);
                root = root->left;
            } else {
                root = s.top();
                res.push_back(root->val);
                s.pop(); 
                root = root->right;
            }
        }

        return res;
    }
};

// 后序
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if (root == nullptr) {
            return {};
        }

        vector<int> res;
        stack<TreeNode*> s;
        while (!s.empty() || root) {
            if (root) {
                s.push(root);
                res.push_back(root->val);
                root = root->right;
            } else {
                root = s.top();
                s.pop();
                root = root->left;
            }
        }

        std::reverse(res.begin(), res.end());
        return res;
    }
};

// 层序
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if (root == nullptr) {
            return res;
        }

        queue<TreeNode*> q;
        q.push(root);
        TreeNode split = TreeNode(INT_MAX);
        q.push(&split);

        vector<int> layer;
        while (!q.empty()) {
            TreeNode *cur = q.front();
            q.pop();

            if (cur == &split) {
                res.push_back(layer);
                layer.clear(); // 注意别忘记清理
                if (!q.empty()) {
                    q.push(&split);
                }
            } else {
                layer.push_back(cur->val);
                if (cur->left) {
                    q.push(cur->left);
                }

                if (cur->right) {
                    q.push(cur->right);
                }
            }
        }

        return res;
    }
};

N叉树

class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> res;
        if (root == nullptr) {
            return res;
        }

        //helper(root, res); // 递归版本
        stack<Node*> s;
        s.push(root);
        while (!s.empty()) {
            Node* cur = s.top();
            s.pop();
            if (cur != nullptr) {
                res.push_back(cur->val);
            }

            std::reverse(cur->children.begin(), cur->children.end());
            for (auto &node : cur->children) {
                s.push(node);
            }
        }
        return res;
    }

    /*void helper(Node* root, vector<int>& out) {
        if (root == nullptr) {
            return;
        }

        out.push_back(root->val);
        for (int i = 0; i < root->children.size(); ++i) {
           helper(root->children[i], out);
        }
    }
    */
};

基础排序

/* 归并排序 */

// merge
void merge(vector<int>& nums, int low, int high, vector<int>& tmp) {
    int mid = (low + high) / 2;
    int leftIndex = low;
    int rightIndex = mid + 1;
    int resLeftIndex = leftIndex;

    while (leftIndex <= mid && rightIndex <= high) {
        // leftIndex 135 
        // rightIndex 246
        // resLeftIndex 123456
        if (nums[leftIndex] >= nums[rightIndex]) {
            tmp[resLeftIndex++] = nums[rightIndex++];
        }
        else {
            tmp[resLeftIndex++] = nums[leftIndex++];
        }
    }

    while (leftIndex <= mid) {
        tmp[resLeftIndex++] = nums[leftIndex++];
    }
    while (rightIndex <= high) {
        tmp[resLeftIndex++] = nums[rightIndex++];
    }

    for (int i = low; i <= high; ++i) { // 易错点 <=
        nums[i] = tmp[i];
    }
}

// 分治
void divideConquer(vector<int>& nums, int low, int high, vector<int>& tmp) {
    if (low >= high) {
        return;
    }

    // 分而治之
    divideConquer(nums, low, (high + low) / 2, tmp);
    divideConquer(nums, (high + low) / 2 + 1, high, tmp);

    // 合并有序数组
    merge(nums, low, high, tmp);
}

void mergeSort(vector<int>& nums) {
    if (nums.empty()) {
        return;
    }

    vector<int> tmp(nums.size());
    divideConquer(nums, 0, nums.size() - 1, tmp);
}

int main()
{
    vector<int> nums = { 2, -1, 4, 55, 0, 67, -23, 5, 9 };

    //quickSortNotR(nums, 0, nums.size() - 1);
    mergeSort(nums);

    for (auto item : nums) {
        cout << item << " ";
    }

    cout << endl;
    return 0;
}

// 快速排序
int getPartSortIndex(vector<int>& nums, int low, int high) {
    
    int tmp = nums[low]; // TODO:随机取值
    int i = low;
    int j = high;
    
    while (i <j) {
        while (i < j && nums[j] >= tmp) {
            j--;
        }

        nums[i] = nums[j];

        while (i < j && nums[i] <= tmp) {
            i++;
        }

        nums[j] = nums[i];
    }

    nums[i] = tmp;

    return i;
}

// 递归版本
void quickSort(vector<int>& nums, int low, int high) {
    if (low >= high) {
        return;
    }

    int index = getPartSortIndex(nums, low, high);
    quickSort(nums, low, index - 1);
    quickSort(nums, index + 1, high);
}

// 非递归版本
void quickSortNotR(vector<int>& nums, int low, int high) { 
    if (low >= high) {
        return;
    }

    stack<int> s;
    s.emplace(low);
    s.emplace(high);
    
    while (!s.empty()) {
        int right = s.top();
        s.pop();

        int left = s.top();
        s.pop();

        int index = getPartSortIndex(nums, left, right);
        if (index - 1 > left) {
            s.emplace(left);
            s.emplace(index - 1);
        }

        if (index + 1 < right) {
            s.emplace(index + 1);
            s.emplace(right);
        }
    }
}

典型常考类型及题目(需要面试前回顾下)

        题目列举

                二分

                链表与数组

                二叉树与分治

                        二叉树翻转——注意:用递归、DFS、栈分别实现下

                排序

                广度优先

                哈希与堆(应急不优先)

                        第K个大或者小元素

                        异位词

                两根指针(应急不优先)

                深度优先(应急不优先)

                动态规划(应急不优先)

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