3Sum Closest -- LeetCode

原题链接: http://oj.leetcode.com/problems/3sum-closest/
这道题跟3Sum很类似，区别就是要维护一个最小的diff，求出和目标最近的三个和。brute force时间复杂度为O(n^3)，优化的解法是使用排序之后夹逼的方法，总的时间复杂度为O(n^2+nlogn)=(n^2),空间复杂度是O(n),代码如下：

public int threeSumClosest(int[] num, int target) {
    if(num == null || num.length<=2)
        return Integer.MIN_VALUE;
    Arrays.sort(num);
    int closest = num[0]+num[1]+num[2]-target;    
    for(int i=0;i<num.length-2;i++)
    {
        int cur = twoSum(num,target-num[i],i+1);
        if(Math.abs(cur)<Math.abs(closest))
            closest = cur; 
    }
    return target+closest;
}
private int twoSum(int[] num, int target, int start)
{
    int closest = num[start]+num[start+1]-target;
    int l = start;
    int r = num.length-1;
    while(l<r)
    {
        if(num[l]+num[r]==target)
            return 0;
        int diff = num[l]+num[r]-target;
        if(Math.abs(diff)<Math.abs(closest))
            closest = diff;    
        if(num[l]+num[r]>target)
        {
            r--;
        }
        else
        {
            l++;
        }
    }
    return closest;
}

这道题具体的考察点可以参见3Sum，稍微变体一下，其实区别不大。此题更加复杂的扩展是4Sum，请参见4Sum -- LeetCode.