Leetcode 16. 3sum closest

@author stormma
@date 2017/11/03

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生命不息，奮鬥不止

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題目

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).

思路分析

其實就是3sum的翻版，只要掌握了上一個題目的做法，這個理所當然可以套上一套題目的做法。

先確定1一個數，two pointer去查找，查找之後和target比較，比較上一次(sum - target) 和這次的(sum - target)的差值（sum表示我們查找的三個數的和）

代碼實現

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums.length == 1) {
            return nums[0];
        }
        Arrays.sort(nums);

        int result = Integer.MAX_VALUE >> 1;

        for (int i = 0; i < nums.length - 2; i++) {
            int low = i + 1, high = nums.length - 1;
            int goal = target - nums[i];
            while (low < high) {
                if (goal == nums[low] + nums[high]) {
                    return target;
                }
                result = Math.abs(goal - nums[low] - nums[high]) > Math.abs(result - target) ? result: nums[i] + nums[low] + nums[high];
                if (goal > nums[low] + nums[high]) {
                    low++;
                } else {
                    high--;
                }
            }
        }
        return result;
    }
}