leetCode刷题(持续更新c++解法)

1.two sum

思路：一次遍历+hash

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> a;
        vector<int> b;
        for(int i = 0; i < nums.size(); i++) {
            a[nums[i]] = i;
        }
        for(int i = 0; i < nums.size(); i++) {
            int x = target - nums[i];
            if(a.count(x) && a[x] != i) {
                b.push_back(i);
                b.push_back(a[x]);
                break;
            }
        }
        return b;
    }
};

167. Two Sum II - Input array is sorted

思路：双指针遍历

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        vector<int> a;
        if(numbers.empty()) {
            return a;
        }
        int l = 0,r = numbers.size()-1;
        while(l < r) {
            if(numbers[l] + numbers[r] < target){
                l++;
            }
            else if(numbers[l] + numbers[r] > target) {
                r--;
            }
            else {
                a.push_back(l+1);
                a.push_back(r+1);
                break;
            }
        }
        return a;
    }
};

15.3sum

//问题：找数组中和为0的三元组

//方法：排序后，扫描数组，转化为2sum问题

//2sum对应一个循环，3sum对应两重循环，4sum对应三重循环

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > a;
        if(nums.size() < 3)
            return a;
        sort(nums.begin(),nums.end()); 
        for(int i = 0; i < (int)nums.size()-2; i++) {
            if(nums[i] > 0) 
                break;
            if(i > 0 && nums[i-1] == nums[i])   //相同则继续找下一个数
                continue;
            int l = i+1,r = (int)nums.size() - 1;
            int target = 0-nums[i];
            while(l < r) {
                if(nums[l] + nums[r] == target) {
                    a.push_back({nums[i],nums[l],nums[r]});
                    while(l < r&&nums[l] == nums[l+1])  //如果相同则继续找下个数
                        l++;
                    while(l < r&&nums[r] == nums[r-1])
                        r--;
                    l++,r--;
                }
                else if(nums[l] + nums[r] < target)
                    l++;
                else
                    r--;
            }
        }
        return a;
    }
};

16.3sum closest

问题：离目标值最近的三数之和

方法：排序后，扫描a[i],后面在用left和right首尾两指针扫描

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        if(nums.size() < 3)
            return 0;
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(closest - target);
        sort(nums.begin(),nums.end());
        for(int i = 0; i < nums.size() - 2; i++) {
            int l = i+1, r = nums.size() - 1;
            while(l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                int x = abs(sum - target);
                if(x < diff) {
                    diff = x;
                    closest = sum;
                }
                if(sum < target) 
                    l++;
                else
                    r--;
            }
        }
        return closest;
    }
};

18.4sum

/*

问题：找与目标值相等的4个数

三重循环即可

用set可避免重复结果

*/

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        if(nums.size() < 4) 
            return vector<vector<int>>();
        set<vector<int>> res;
        sort(nums.begin(),nums.end());
        for(int i = 0; i < nums.size()-3; i++) {
            for(int j = i+1; j < nums.size() - 2; j++) {
                int l = j + 1,r = nums.size() - 1;
                while(l < r) {
                    int sum = nums[i] + nums[j] + nums[l] + nums[r];
                    if(sum == target) {
                        vector<int> s{nums[i],nums[j],nums[l],nums[r]};
                        res.insert(s);  //set有重复时插入不进去
                        l++,r--;
                    }
                    else if(sum > target) {
                        r--;
                    }
                    else{
                        l++;
                    }
                }
            }
        }
        return vector<vector<int>> (res.begin(),res.end());  //讲set转为vector输出
    }
};

217. Contains Duplicate

问题：判断一个数组是否含重复元素

方法：用map统计每个元素出现的次数

此题比较简单，方法很多

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_map<int,int> dict;
        for(int i:nums) {
            dict[i]++;
            if(dict[i] > 1) {
                return true;
            }
        }
        return false;
    }
};

26. Remove Duplicates from Sorted Array

问题：去除有序序列中的重复数字

方法一：双指针法(覆盖法)

当a[i] != a[index-1] 用a[i]覆盖a[index]

相等时不覆盖，不等时覆盖，index代表了新数组的索引，i代表了旧数组索引,将无重复数依次移动到前面

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.empty()) {
            return 0;
        }
        int index = 1;
        for(int i = 0; i < nums.size(); i++) {
            if(nums[i] != nums[index-1]){
                nums[index] = nums[i];
                index++;
            }
        }
        return index;
    }
};