16. 3Sum Closest

题目：

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

链接： http://leetcode.com/problems/3sum-closest/

题解：

这道题跟3 sum很像，也是先对数组进行排序，然后通过3指针法对结果进行夹逼。

Time Complexity - O(n²)，Space Complexity - O(1)。

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if(nums == null || nums.length < 3)
            return 0;
        Arrays.sort(nums);
        int left, right, result = 0, diff = Integer.MAX_VALUE;
        
        for(int i = 0; i < nums.length - 2; i ++){
            left = i + 1;
            right = nums.length - 1;
            
            while(left < right){
                int sum = nums[i] + nums[left] + nums[right];
                if(sum == target)
                    return sum;
                else if(sum < target)
                    left ++;
                else
                    right --;
                if(Math.abs(sum - target) < diff){
                    diff = Math.abs(sum - target);
                    result = sum;
                }
            }
        }
        
        return result;
    }
}

 

二刷：

代码基本和3Sum一样，时间复杂度也一样。关键点还是去重的pruning，去重能节约不少时间。

Java:

Time Complexity - O(n²)， Space Complexity - O(1)

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3) {
            return 0;
        }
        Arrays.sort(nums);
        int diff = Integer.MAX_VALUE, result = 0;
        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;
                } else if (sum < target) {
                    lo++;
                    while (lo < hi && nums[lo] == nums[lo - 1]) {
                        lo++;
                    }
                } else {
                    hi--;
                    while (lo < hi && nums[hi] == nums[hi + 1]) {
                        hi--;
                    }
                }
                if (Math.abs(sum - target) < diff) {
                    diff = Math.abs(sum - target);
                    result = sum;
                }
            }
        }
        return result;
    }
}

 

Python: 

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        if not nums or len(nums) < 3:
            return 0
        diff = 1 << 32
        res = 0
        nums.sort()
        for i in range(0, 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
                elif sum < target:
                    lo += 1
                    while lo < hi and nums[lo] == nums[lo - 1]:
                        lo += 1
                else:
                    hi -= 1
                    while lo < hi and nums[hi] == nums[hi + 1]:
                        hi -= 1
                if abs(sum - target) < diff:
                    diff = abs(sum - target)
                    res = sum
        return res        
                
        
                    

 

三刷：

先对数组进行排序，初始化closest = 0，diff = Integer.MAX_VALUE, 然后遍历数组。注意要去重复，并且在计算完每个sum以后，要尝试更新closest。 最后返回结果。去重复包括遍历i的时候，以及lo++和hi--的时候。

Java:

Time Complexity - O(n²)，Space Complexity - O(1)。

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3) return 0;
        Arrays.sort(nums);
        int closest = 0, diff = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int lo = i + 1, hi = nums.length - 1;
            int sum = 0;
            while (lo < hi) {
                sum = nums[i] + nums[lo] + nums[hi];
                if (sum == target) {
                    return sum;
                } else if (sum < target) {
                    lo++;
                    while (lo < hi && nums[lo] == nums[lo - 1]) lo++;    
                } else {
                    hi--;
                    while (lo < hi && nums[hi] == nums[hi + 1]) hi--;
                }
                
                if (Math.abs(target - sum) < diff) {
                    diff = Math.abs(target - sum);
                    closest = sum;
                }
            }
        }
        return closest;
    }
}

 

Reference: