1. 两数之和

直接利用hashmap存储值和对于索引，利用target-nums[i]去哈希表里找对应数值。返回下标。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> mp;
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            if (mp.find(target - nums[i]) != mp.end()) {
                res.push_back(i);
                res.push_back(mp[target - nums[i]]);
            }
            mp[nums[i]] = i;
        }
        return res;
    }
};

15. 三数之和

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        if (nums.empty() || nums.size() < 3) {
            return res;
        } 

        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] > 0) {
                break;
            }

            if (i > 0 && nums[i] == nums[i-1]) {
                continue;
            }

            int left = i + 1;
            int right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum > 0) {
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    res.push_back({nums[i], nums[left], nums[right]});
                    while (left < right && nums[left] == nums[left+1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right-1]) {
                        right--;
                    }
                    left++;
                    right--;
                }
            }
        }
        return res;
    }
};

16. 最接近的三数之和

排序+双指针

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        long res = INT_MAX;

        for (int i = 0; i < nums.size(); ++i) {
            int left = i + 1;
            int right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (abs(target - sum) < abs(target - res)) {
                    res = sum;
                }

                if (sum > target) {
                    right--;
                } else if (sum < target) {
                    left++;
                } else {
                    return res;
                }
            }
        }
        return res;
    }
};

 18. 四数之和

本题与「15. 三数之和」相似，解法也相似。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> quadruplets;
        if (nums.size() < 4) {
            return quadruplets;
        }
        sort(nums.begin(), nums.end());
        int length = nums.size();
        for (int i = 0; i < length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }
            if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {
                continue;
            }
            for (int j = i + 1; j < length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }
                if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {
                    continue;
                }
                int left = j + 1, right = length - 1;
                while (left < right) {
                    long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        quadruplets.push_back({nums[i], nums[j], nums[left], nums[right]});
                        while (left < right && nums[left] == nums[left + 1]) {
                            left++;
                        }
                        left++;
                        while (left < right && nums[right] == nums[right - 1]) {
                            right--;
                        }
                        right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return quadruplets;
    }
};

560. 和为 K 的子数组

使用前缀和

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        mp[0] = 1;
        int sum = 0, count = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            if (mp.find(sum - k) != mp.end()) {
                count += mp[sum - k];
            }
            mp[sum]++;
        }
        return count;
    }
};

53. 最大子数组和

动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        vector<int> dp(nums.size());
        dp[0] = nums[0];
        
        for (int i = 1; i < nums.size(); ++i) {
            dp[i] = max(dp[i-1] + nums[i], nums[i]);
        }

        int res = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            res = max(res, dp[i]);
        }

        return res;
    }
};

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int pre = 0;
        int res = nums[0];
        
        for (int i = 0; i < nums.size(); ++i) {
            pre = max(pre + nums[i], nums[i]);
            res = max(res, pre);
        }

        return res;
    }
};