算法刷题计划（十六）丢失的数字、范围求和 II、外观数列

一、丢失的数据

• 题目
  

---

• 题解
  方法一（数学整数相减）：

class Solution {
     
public:
    int missingNumber(vector<int>& v) {
     
        int n=v.size();
        int sum=n*(n+1)/2;
        int total=accumulate(v.begin(),v.end(),0);
        return sum-total;
    }
};

方法二（遍历）：

class Solution {
     
public:
    int missingNumber(vector<int>& nums) {
     
        int n = nums.size(),i=0;
        sort(nums.begin(),nums.end());
        while(n>0){
     
            if(nums[i]==i)  n--,i++;
            else{
     
                break;
            }
        }
        return i;
    }
};

方法三（哈希）：

class Solution {
     
public:
    int missingNumber(vector<int>& v) {
     
    unordered_set<int>us;
    int n =v.size();
    for(auto&s:v){
     
        us.insert(s);
    }
    int k=-1;
    for(int i=0;i<=n;i++){
     
        if(!us.count(i)){
     
            k=i;
            break;
        }
        // cout<<*us.find(i);
    }
    return k;
    }
};

方法四（位运算）：

class Solution {
     
public:
    int missingNumber(vector<int>& nums) {
     
        int res = 0;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
     
            res ^= nums[i];
        }
        for (int i = 0; i <= n; i++) {
     
            res ^= i;
        }
        return res;
    }
};

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范围求和 II

• 题目：
  
  

---

• 题解

class Solution {
     
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
     
        int mina = m, minb = n;
        for (const auto& op: ops) {
     
            mina = min(mina, op[0]);
            minb = min(minb, op[1]);
        }
        return mina * minb;
    }
};

外观数列

• 题目：
  
  

---

• 题解：

class Solution {
     
public:
     string countAndSay(int n) {
     
         if(n==1)    return "1";
         string strCallBack=countAndSay(n-1);
         int times=0;
         char s=strCallBack[0];
         string callBack;
         for(int i=0;i<strCallBack.size();i++){
     
             if(strCallBack[i]==s){
     
                 times++;
           }else{
     
                callBack+=to_string(times)+strCallBack[i-1];
                times=1;
                 s=strCallBack[i];
            }
             if(i==strCallBack.size()-1){
     
                callBack+=to_string(times)+strCallBack[i];
            }
        }
         return callBack;
     }

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组合总和

• 题目
  

---

• 题解

class Solution {
     
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& candidates, int target, int sum, int startIndex) {
     
        if (sum == target) {
     
            result.push_back(path);
            return;
        }

        // 如果 sum + candidates[i] > target 就终止遍历
        for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
     
            sum += candidates[i];
            path.push_back(candidates[i]);
            backtracking(candidates, target, sum, i);
            sum -= candidates[i];
            path.pop_back();

        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
     
        result.clear();
        path.clear();
        sort(candidates.begin(), candidates.end()); // 需要排序
        backtracking(candidates, target, 0, 0);
        return result;
    }
};