Job Interview Notes: Algorithms

ToC


记录一些刷到的算法题的解法。

P1. two sum

很简单的一道题但是发现在做的时候其实有一些需要注意的地方

  1. unordered_map instead of map,因为map底层是红黑树,查询效率是O(logn),而unordered_map底层是哈希表,查询效率是O(1)(worst case O(n))
  2. unordered_map[key]如果key不存在,会返回0,但是有情况是0是有效值,所以需要用find判断是否存在
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        std::unordered_map<int, int> m;
        for (int i = 0, n = nums.size(); i < n; i++) {
            if (auto p = m.find(target - nums[i]); p != m.end()) {
                return {i, p->second};
            }
            m[nums[i]] = i;
        }
        return {-1, -1};
    }
};

P2. longest consecutive sequence

两个特点:

  1. 如果\(x-1\)存在,那么\(x\)一定不是最长连续序列的起点,跳过
  2. 如果\(x-1\)不存在,那么\(x\)一定是最长连续序列的起点,开始向后查询连续序列的长度,更新最大值
class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end()); // use a set
        int ans = 0;

        for (auto num : s) {
            // if we have (num - 1) in the set,
            // num will not give us the answer.
            if (s.find(num - 1) != s.end()) {
                continue;
            }

            // this inner loop will run only n times,
            // which gives us a time complexity of O(n)
            int i = 1;
            while (s.find(num + i) != s.end())
                i++;
            ans = std::max(ans, i);
        }

        return ans;
    }
};

P3. container with most water

把下标\(i, j\)围成的容器的盛水量设为\(m(i, j)\),有

\[m(i, j) = \min(x_i, x_j) * (j - i)\]

有(\(x_1, x_2, ..., x_{n-1}, x_n\)),目前双指针位于\(x_1\)和\(x_n\),设\(x_1 < x_n\),那么有

\[m(1,n) = x_1 * (n - 1)\]

同时

\[m(1, n-1) = \min(x_1, x_{n-1}) * (n - 2)\] \[\le x_1 * (n - 2)\] \[< x_1 * (n - 1)\] \[< m(1, n)\]

所以每次都应该移动较小的那个指针,直到两个指针相遇。

class Solution {
public:
    int maxArea(vector<int>& height) {
        int i = 0, j = height.size() - 1;
        int ans = 0;
        while (i < j) {
            int current = std::min(height[i], height[j]) * (j - i);
            ans = std::max(ans, current);
            if (height[i] < height[j]) {
                i++;
            } else {
                j--;
            }
        }
        return ans;
    }
};

P4. 3sum

  1. 三重循环,排序之后,枚举\(a\),在\(a\)之后枚举\(b\)和\(c\),使用双指针优化
  2. 需要去重,跳过重复的\(a\),\(b\)即可
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        std::sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            // duplicated
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }

            // don't put this inside the loop below
            // otherwise it's not a two pointers but
            // a time complexity of O(n3)
            int k = n - 1;

            for (int j = i + 1; j < n; j++) {
                // duplicated
                if (j != i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }

                while (j < k && nums[i] + nums[j] + nums[k] > 0) {
                    k--;
                }

                // all following j will not satisfy
                if (j == k) {
                    break;
                }

                if (nums[i] + nums[j] + nums[k] == 0) {
                    ans.push_back({nums[i], nums[j], nums[k]});
                }
            }
        }
        return ans;
    }
};

P5. trapping rain water

class Solution {
public:
    int trap(vector<int>& height) {
        vector<int> forward(height.size());
        vector<int> backward(height.size());
        int ans = 0;

        int m = 0;
        // 正向计算每个下标处最多能接多少水
        for (int i = 0; i < height.size(); i++) {
            m = std::max(m, height[i]);
            forward[i] = m - height[i];
        }

        m = 0;
        // 反向计算每个下标处最多能接多少水
        for (int i = height.size() - 1; i >= 0; i--) {
            m = std::max(m, height[i]);
            backward[i] = m - height[i];

            // 最终答案是每个下标处按两种方式计算后最多能接水的最小值
            ans += min(backward[i], forward[i]);
        }
        return ans;
    }
};

P6. longest substring without repeating characters

P7. median of two sorted arrays

P8. find all anagrams in a string

  1. 对每个字母进行计数,然后记录当前窗口中字母数量和p中字母数量不同的字母的数量,如果不同的字母数量为0,那么说明当前窗口是一个anagram,记录答案。
  2. 注意在移动窗口时,如果移出的和移入的字母是相同的,那么判断值是否为0会被判定两次,因此需要分别判断(代码的注释)
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        unordered_map<char, int> counts;
        vector<int> ans;
        int diff = 0;

        if (p.size() > s.size()) {
            return {};
        }

        for (int i = 0; i < p.size(); i++) {
            counts[s[i]]++;
            counts[p[i]]--;
        }

        for (int i = 'a'; i <= 'z'; i++) {
            if (counts[i] != 0) {
                diff += 1;
            }
        }

        if (diff == 0) {
            ans.emplace_back(0);
        }

        for (int i = p.size(); i < s.size(); i++) {
            counts[s[i]]++;
            if (counts[s[i]] == 0) {
                diff -= 1;
            } else if (counts[s[i]] == 1) {
                diff += 1;
            }

            // 不能写在循环开头,因为删掉的和新增的可能是同一个字符
            counts[s[i - p.size()]]--;
            if (counts[s[i - p.size()]] == 0) {
                diff -= 1;
            } else if (counts[s[i - p.size()]] == -1) {
                diff += 1;
            }

            if (!diff) {
                ans.emplace_back(i - p.size() + 1);
            }
        }

        return ans;
    }
};

P9. subarray sum equals k

刚开始想用双指针,但是写出来问题比较多,看题解才看到前缀和+哈希的解法,前缀和也不需要存数组,只需要一个变量即可。

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int pre = 0;
        unordered_map<int, int> m;
        m[0] = 1;
        int ans = 0;
        for (auto num : nums) {
            pre += num;
            ans += m[pre - k];
            m[pre] += 1;
        }
        return ans;
    }
};

P10. reverse nodes in k-group

  1. 每次翻转k个节点,递归剩下的节点
  2. 或者,使用一个额外数组存储然后翻转

P11. edit distance

  1. 对于两个字符串\(s_1\)和\(s_2\),分别有对字符串\(s_1\)或\(s_2\)进行插入/删除/替换操作,一共是6种情况,而实际上对\(s_1\)进行删除操作就等价于对\(s_2\)进行插入操作,因此最后的情况是:对\(s_1\)进行插入操作,对\(s_2\)进行插入操作,对\(s_1\)进行替换操作。一共三种。
  2. 对于字符串如”adadh”和”dah”,以及”adadh”和”da”,前者至多比后者操作次数多一次。
  3. 也就是说,对于字符串\(s_1\)的前\(i\)个字符和字符串\(s_2\)的前\(j\)个字符,编辑距离最多是前\(i\)个字符和前\(j-1\)个字符的编辑距离+1,前\(i-1\)个字符和前\(j\)个字符的编辑距离+1,前\(i-1\)个字符和前\(j-1\)个字符的编辑距离+1,取最小值。 4.

[P12. ]

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first published: 2024-09-22 16:23:43 CST
last modified: 2024-11-16 16:28:52 CST

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