代码随想录算法训练营第七天| 454.四数相加II、383. 赎金信、15. 三数之和、18. 四数之和

454.四数相加II

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文章讲解

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        unordered_map<int, int> rec;
        int i, j, size = nums1.size(), ans = 0;
        for (i = 0; i < size; i++) {
            for (j = 0; j < size; j++) {
                rec[nums1[i]+nums2[j]]++;
            }
        }
        for(i = 0; i < size; i++) {
            for (j = 0; j < size; j++) {
                auto it = rec.find(-nums3[i]-nums4[j]);
                if (it != rec.end()) ans += it->second;
            }
        }
        return ans;
    }
};

和 “1. 两数之和” 相似，将元素保存在key中，将次数或下标保存在value中

383. 赎金信

力扣题目链接
文章讲解

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int rec[26] = {0};
        for (char c : magazine) rec[c - 'a']++;
        for (char c : ransomNote) 
            if (--rec[c - 'a'] < 0) return false;
        return true;
    }
};

使用数组记录方便些

15. 三数之和

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文章讲解

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        int i, left, right, size = nums.size();
        for (i = 0; i < size - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            left = i + 1, right = size - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum > 0) {
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    ans.push_back({nums[i], nums[left], nums[right]});
                    while (left < right && nums[right] == nums[right - 1]) right--;
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    right--, left++;
                }
            }
        }
        return ans;
    }
};

去重和去重时机是关键

18. 四数之和

力扣题目链接
文章讲解

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        int i, j, left, right, size = nums.size();
        for (i = 0; i < size - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            for (j = i + 1; j < size - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                left = j + 1, right = size - 1;
                while (left < right) {
                    long long sum = (long long) nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum > target) {
                        right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        ans.push_back({nums[i], nums[j], nums[left], nums[right]});
                        while (left < right && nums[right] == nums[right - 1]) right--;
                        while (left < right && nums[left] == nums[left + 1]) left++;
                        left++, right--;
                    }
                }
            }
        }
        return ans;
    }
};

思路和上一题一样，要明白双指针的作用