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 + num + num - target > 0 && k > (j+1)) k--; int close = Math.abs(num + num + num - target); if (close < tempClose){ tempClose = close; tempsum = num + num + num; } if (k < num.length - 1){ close = Math.abs(num + num + num - target); if (close < tempClose){ tempClose = close; tempsum = num + num + num; } k++; } } } return tempsum; }}
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