[Leetcode] 016. 3Sum Closest (Medium) (C++/java/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

016.3sum_closest (Medium) links：

Title: https://oj.leetcode.com/problems/3sum-closest/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Finds three numbers in a given sequence, making and closest to target.

Analysis：

With 015. 3Sum (Medium) is similar, even simpler.
Or the first sort, and then left and right clamping force.

Code：

C++:

Class Solution {Public:int threesumclosest (vector<int> &num, int target) {int ret = num[0] + num[1] + num[2];int Len = Num.size (), Sort (Num.begin (), Num.end ()), for (int i = 0; I <= len-3; i++) {//First Number:num[i]int j = i + 1;//second Numberint k = len-1;//Third Numberwhile (J < k) {int sum = Num[i] + num[j] + num[k];if (ABS (Sum-target ) < ABS (ret-target)) ret = sum;if (sum < target) {++j;} else if (Sum > Target) {--k;} else {++j;--k;}}} return ret;}};

Java:

public class Solution {public    int threesumclosest (int[] num, int target) {        arrays.sort (num);        int ret = num[0] + num[1] + num[2];        int len = num.length;        for (int i = 0; I <= len-3; i++) {            //First number:num[i]            int j = i + 1;//second number            int k = len- 1;//Third Number            while (J < k) {                int sum = num[i] + num[j] + num[k];                if (Math.Abs (Sum-target) < Math.Abs (ret-target))                    ret = sum;                if (sum < target) {                    ++j;                } else if (Sum > Target) {                    --k;                } else {                    ++j;                    --k        ;        }}} return ret;    }}

Python:

Class solution:    # @return An integer    def threesumclosest (self, num, target):        if not len (num):            return 0< C4/>ret = num[0] + num[1] + num[2]        num.sort () for        I in range (len (num)-2):            j = i + 1            k = len (num)-1
   
    while J < k:                Tsum = num[i] + num[j] + num[k]                if ABS (Tsum-target) < ABS (Ret-target):                    ret = tsum
    if Tsum < target:                    j + = 1                elif tsum > Target:                    k-= 1                Else:                    j + = 1                    k-= 1        retur n RET
   

[Leetcode] 016. 3Sum Closest (Medium) (C++/java/python)