LeetCode 刷题记录 16. 3Sum Closest

最新推荐文章于 2021-02-05 15:36:02 发布

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本文介绍了一种解决三数之和最接近目标值问题的算法，该问题要求在给定数组中找到三个整数，使它们的和最接近给定的目标值。通过排序和双指针技术，算法有效地找到了最优解，并提供了C++、Java和Python三种语言的实现代码。

题目：
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
解法：
解法和3Sum相似，不过要记录与目标值的差值，每次差值变小时更新结果
c++：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int res = 0;
        int min_gap = INT_MAX;
        sort(nums.begin(), nums.end());
       
        
        for(int i = 0; i < nums.size() - 2; i++){
            
           if(i > 0 && nums[i] == nums[i - 1]) continue;
            int lo = i + 1;
            int hi = nums.size() - 1;
            
            while(lo < hi){
                int sum = (nums[i] + nums[lo] + nums[hi]);
                if(sum == target) return sum;
                if(abs(sum - target) < min_gap){
                    res = sum;
                    min_gap = abs(sum - target);
                }
                
                    
                if(sum < target){
                   while(lo < hi && nums[lo] == nums[lo + 1]) lo++;
                    lo++;
                } else {
                    while(lo < hi && nums[hi] == nums[hi - 1]) hi--;
                    hi--;
                }
            }
            
            
        }
        return res;
    }
};

java：

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int res = 0;
        int min_gap = Integer.MAX_VALUE;
        Arrays.sort(nums);
       
        
        for(int i = 0; i < nums.length - 2; i++){
            
           if(i > 0 && nums[i] == nums[i - 1]) continue;
            int lo = i + 1;
            int hi = nums.length - 1;
            
            while(lo < hi){
                int sum = (nums[i] + nums[lo] + nums[hi]);
                if(sum == target) return sum;
                if(Math.abs(sum - target) < min_gap){
                    res = sum;
                    min_gap = Math.abs(sum - target);
                }
                
                    
                if(sum < target){
                   while(lo < hi && nums[lo] == nums[lo + 1]) lo++;
                    lo++;
                } else {
                    while(lo < hi && nums[hi] == nums[hi - 1]) hi--;
                    hi--;
                }
            }
            
            
        }
        return res;
    }
}

python:

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        res = 0
        min_gap = sys.maxsize
        nums.sort()
       
        
        for i in range(len(nums) - 2):
            
            if i > 0 and nums[i] == nums[i - 1]: continue
            lo = i + 1
            hi = len(nums) - 1
            
            while lo < hi:
                sum = (nums[i] + nums[lo] + nums[hi])
                if sum == target: return sum
                if abs(sum - target) < min_gap:
                    res = sum
                    min_gap = abs(sum - target)
                
                
                    
                if sum < target:
                    #while lo < hi and nums[lo] == nums[lo + 1]: lo += 1
                    lo += 1
                else:
                    #while lo < hi and nums[hi] == nums[hi - 1]: hi -= 1
                    hi -= 1
               
            
        
        return res;