n数之和

n 数之和

本质上就是将双指针作为等效循环（n 层循环 + 双指针 = n + 1 层循环），实现复杂度降维

两数之和（核心）

排序 + 首尾双指针

vector<vector<int>> twoSum(vector<int> &nums, int target) {
    vector<vector<int>> res;
    sort(nums.begin(), nums.end());
    int left = 0, right = nums.size() - 1;
    //要求两数之和不能取同一个数，故终止条件为 left < right
    while (left < right) {
        int sum = nums[left] + nums[right];
        if (sum == target) {
            res.push_back({nums[left], nums[right]});
            // 去重
            while (left < right && nums[left] == nums[left + 1]) left++;
            while (left < right && nums[right] == nums[right - 1]) right--;
            left++;
            right--;
        } else if (sum > 0) {
            //大了，right 左移
            right--;
        } else {
            //小了，left 右移
            left++;
        }
    }
    return res;
}

三数之和

vector<vector<int>> threeSum(vector<int> &nums, int target) {
    vector<vector<int>> res;
    sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size(); i++) {
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }
        //内部复用两数之和的逻辑，但是左指针为外层遍历元素的下一个元素
        int left = i + 1, right = nums.size() - 1;
        while (left < right) {
            //符合要求，处理结果
            if (nums[i] + nums[left] + nums[right] == target) {
                //获取第一个三元组后再做剪枝，防止漏掉[0, 0, 0]的情况
                res.push_back({nums[i], nums[left], nums[right]});
                // 去重
                while (left < right && nums[left] == nums[left + 1]) left++;
                while (left < right && nums[right] == nums[right - 1]) right--;
                left++;
                right--;
            } else if (nums[i] + nums[left] + nums[right] > target) {
                //大了，right 左移
                right--;
            } else {
                //小了，left 右移
                left++;
            }
        }
    }
    return res;
}

四数之和

注意防止求和溢出

vector<vector<int>> fourSum(vector<int>& nums, int target) {
    vector<vector<int>> result;
    sort(nums.begin(), nums.end());
    for (int k = 0; k < nums.size(); k++) {
        // 对nums[k]去重
        if (k > 0 && nums[k] == nums[k - 1]) {
            continue;
        }
        for (int i = k + 1; i < nums.size(); i++) {
            // 对nums[i]去重
            if (i > k + 1 && nums[i] == nums[i - 1]) {
                continue;
            }
            int left = i + 1;
            int right = nums.size() - 1;
            while (right > left) {
                long sum = (long) nums[k] + nums[i] + nums[left] + nums[right];
                if (sum > target) {
                    right--;
                } else if (sum  < target) {
                    left++;
                } else {
                    result.push_back(vector<int>{nums[k], nums[i], nums[left], nums[right]});
                    while (right > left && nums[right] == nums[right - 1]) right--;
                    while (right > left && nums[left] == nums[left + 1]) left++;
                    right--;
                    left++;
                }
            }

        }
    }
    return result;
}

n 数之和（推广）

在外面再套一层 for 循环即可，注意控制非最外层循环起点为 上一层循环变量 + 1

//for 循环参考
for (int k = 0; k < nums.size(); k++) {
    // 对nums[k]去重
    if (k > 0 && nums[k] == nums[k - 1]) {
        continue;
    }
}

代码汇总

#include <iostream>

using namespace std;

//region 规范化输出
void printVector(vector<int> &nums) {
    cout << "[";
    for (int i = 0; i < nums.size(); i++) {
        cout << nums[i];
        if (i != nums.size() - 1) {
            cout << ", ";
        }
    }
    cout << "]";
}

void printResult(vector<vector<int>> &res) {
    cout << "[";
    for (int i = 0; i < res.size(); i++) {
        printVector(res[i]);
        if (i != res.size() - 1) {
            cout << ", ";
        }
    }
    cout << "]";
}
//endregion

vector<vector<int>> twoSum(vector<int> &nums, int target) {
    vector<vector<int>> res;
    sort(nums.begin(), nums.end());
    int left = 0, right = nums.size() - 1;
    //要求两数之和不能取同一个数，故终止条件为 left < right
    while (left < right) {
        int sum = nums[left] + nums[right];
        if (sum == target) {
            res.push_back({nums[left], nums[right]});
            // 去重
            while (left < right && nums[left] == nums[left + 1]) left++;
            while (left < right && nums[right] == nums[right - 1]) right--;
            left++;
            right--;
        } else if (sum > 0) {
            //大了，right 左移
            right--;
        } else {
            //小了，left 右移
            left++;
        }
    }
    return res;
}

vector<vector<int>> threeSum(vector<int> &nums, int target) {
    vector<vector<int>> res;
    sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size(); i++) {
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }
        //内部复用两数之和的逻辑，但是左指针为外层遍历元素的下一个元素
        int left = i + 1, right = nums.size() - 1;
        while (left < right) {
            //符合要求，处理结果
            if (nums[i] + nums[left] + nums[right] == target) {
                //获取第一个三元组后再做剪枝，防止漏掉[0, 0, 0]的情况
                res.push_back({nums[i], nums[left], nums[right]});
                // 去重
                while (left < right && nums[left] == nums[left + 1]) left++;
                while (left < right && nums[right] == nums[right - 1]) right--;
                left++;
                right--;
            } else if (nums[i] + nums[left] + nums[right] > target) {
                //大了，right 左移
                right--;
            } else {
                //小了，left 右移
                left++;
            }
        }
    }
    return res;
}

vector<vector<int>> fourSum(vector<int>& nums, int target) {
    vector<vector<int>> result;
    sort(nums.begin(), nums.end());
    for (int k = 0; k < nums.size(); k++) {
        // 对nums[k]去重
        if (k > 0 && nums[k] == nums[k - 1]) {
            continue;
        }
        for (int i = k + 1; i < nums.size(); i++) {
            // 对nums[i]去重
            if (i > k + 1 && nums[i] == nums[i - 1]) {
                continue;
            }
            int left = i + 1;
            int right = nums.size() - 1;
            while (right > left) {
                long sum = (long) nums[k] + nums[i] + nums[left] + nums[right];
                if (sum > target) {
                    right--;
                } else if (sum  < target) {
                    left++;
                } else {
                    result.push_back(vector<int>{nums[k], nums[i], nums[left], nums[right]});
                    while (right > left && nums[right] == nums[right - 1]) right--;
                    while (right > left && nums[left] == nums[left + 1]) left++;
                    right--;
                    left++;
                }
            }

        }
    }
    return result;
}

int main() {
    vector<int> nums = {-1, 0, 1, 2, -1, -4};
    printVector(nums);
    cout << endl;

    int target = 0;
    vector<vector<int>> res = twoSum(nums, target);
    printResult(res);
    cout << endl;

    res = threeSum(nums, target);
    printResult(res);
    cout << endl;

    res = fourSum(nums, target);
    printResult(res);
    cout << endl;

    return 0;
}