3Sum Closest

这题的思路和3Sum一样，只是要考虑一下k在j+1之后是不是需要减1，因为不是0而是绝对值最小，所以有点难考虑= = 我就懒得去想直接没-1 汗啊

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        Arrays.sort(num);
        int tempClose = Integer.MAX_VALUE;
        int tempsum = Integer.MAX_VALUE;
        for (int i = 0; i != num.length; ++i){            
            int k = num.length - 1;
            for (int j = i + 1; j < k; j++){
                while (num[i] + num[j] + num[k] - target > 0 && k > (j+1))
                    k--;
                int close = Math.abs(num[i] + num[j] + num[k] - target);
                if (close < tempClose){
                    tempClose = close;
                    tempsum = num[i] + num[j] + num[k];
                }
                if (k < num.length - 1){
                    close = Math.abs(num[i] + num[j] + num[k+1] - target);
                    if (close < tempClose){
                        tempClose = close;
                        tempsum = num[i] + num[j] + num[k+1];
                    }
                    k++;
                }
            }
        }
        return tempsum;        
    }
}