算法题：16. 最接近的三数之和（Python & Java 详解）

 

 解题思路

Step1：先对数组排序，然后设置3个指针，指针1遍历范围为（0~数组长度减2）。

Step2：指针1位置确定时，指针1后面的数组元素首位各放置一个指针（指针2、指针3）。

Step3：如果三数之和=target，则返回target值；如果三数之和指针2往后移动，如果三数之和>target，则将指针3往前移动。

Step4：当指针2和指针3重合时，则将指针1往后移动。

Step5：重复 Step2 到 Step4。直到指针1遍历完。

Java代码

import java.util.Arrays;


public class ThreeSumClosest {
    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.threeSumClosest(new int[]{-1, 2, 1, -4}, 1));
    }
}

class Solution {
    public int threeSumClosest(int[] nums, int target) {

        Arrays.sort(nums);
        int n = nums.length;
        int min_diff = Integer.MAX_VALUE;//当前找到的三数之和与target的最小差值
        int ans = 0;//最小差值对应的三数之和

        int cur_sum = 0;
        int cur_diff = 0;
        for (int i = 0; i < n - 2; i++) {
            //优化一：如果有重复元素则跳过
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            //优化二：如果加上最后面两个数依旧没有target大，则判断一下三数之和之后跳过当前循环
            cur_sum = nums[i] + nums[n - 2] + nums[n - 1];
            if (target >= cur_sum) {
                cur_diff = target - cur_sum;
                if (cur_diff == 0) {
                    return target;
                }
                if (cur_diff < min_diff) {
                    ans = cur_sum;
                    min_diff = cur_diff;
                }
                continue;
            }
            //优化三：前面3个数的和已经>=target，则判断三数之和之后跳出循环
            cur_sum = nums[i] + nums[i + 1] + nums[i + 2];
            if (target <= cur_sum) {
                cur_diff = cur_sum - target;
                if (cur_diff == 0) {
                    return target;
                }
                if (cur_diff < min_diff) {
                    ans = cur_sum;
                }
                break;
            }

            int j = i + 1, k = n - 1;
            while(j < k){
                cur_sum = nums[i] + nums[j] + nums[k];
                if (cur_sum == target) {
                    return target;
                }
                cur_diff = target - cur_sum;
                if (cur_sum < target) {
                    if (cur_diff < min_diff) {
                        ans = cur_sum;
                        min_diff = cur_diff;
                    }
                    j++;
                } else {  //cur_sum > target
                    cur_diff = -cur_diff;
                    if (cur_diff < min_diff) {
                        ans = cur_sum;
                        min_diff = cur_diff;
                    }
                    k--;
                }
            }
        }
        return ans;
    }
}

Python代码

class Solution(object):
    def threeSumClosest(self, nums, target):
        nums = sorted(nums)  # or nums.sort()
        n = len(nums)
        min_diff = float('inf')
        ans = 0
        for i in range(n-2):
            # 优化一：如果有重复元素则跳过
            if i > 0 and nums[i] == nums[i-1]:
                continue
            # 优化二：如果加上最后面两个数依旧没有target大，则判断一下三数之和之后跳过当前循环
            cur_sum = nums[i] + nums[n - 2] + nums[n - 1]
            if target >= cur_sum:
                cur_diff = target - cur_sum
                if cur_diff == 0:
                    return target
                if cur_diff < min_diff:
                    ans = cur_sum
                    min_diff = cur_diff
                continue
            # 优化三：前面3个数的和已经>=target，则判断三数之和之后跳出循环
            cur_sum = nums[i] + nums[i + 1] + nums[i + 2]
            if target <= cur_sum:
                cur_diff = cur_sum - target
                if cur_diff == 0:
                    return target
                if cur_diff < min_diff:
                    ans = cur_sum
                break
            j, k = i + 1, n - 1
            while j < k:
                cur_sum = nums[i] + nums[j] + nums[k]
                if cur_sum == target:
                    return target
                cur_diff = target - cur_sum
                if cur_sum < target:
                    if cur_diff < min_diff:
                        ans = cur_sum
                        min_diff = cur_diff
                    j += 1
                else:
                    cur_diff = -cur_diff
                    if cur_diff < min_diff:
                        ans = cur_sum
                        min_diff = cur_diff
                    k -= 1
        return ans

if __name__ == '__main__':
    sol = Solution()
    print(sol.threeSumClosest([-1, 2, 1, -4], 1))  # 2

完整题目

16. 最接近的三数之和

给你一个长度为 n 的整数数组 nums 和 一个目标值 target。请你从 nums 中选出三个整数，使它们的和与 target 最接近。

返回这三个数的和。

假定每组输入只存在恰好一个解。

示例 1：

输入：nums = [-1,2,1,-4], target = 1
输出：2
解释：与 target 最接近的和是 2 (-1 + 2 + 1 = 2) 。

示例 2：

输入：nums = [0,0,0], target = 1
输出：0

提示：

• 3 <= nums.length <= 1000
• -1000 <= nums[i] <= 1000
• -10^4 <= target <= 10^4