LC.P16[最接近的三数之和]

方法一：排序+双指针

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class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int n = nums.length, ans = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < n - 2; ++i) {
            int j = i + 1, k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == target) return sum;
                if (Math.abs(sum - target) < Math.abs(ans - target)) ans = sum;
                if (sum < target) ++j;
                else if (sum > target) --k;
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n^2)$
  • 空间复杂度：$O(logn)$

方法二：优化

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class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int n = nums.length, ans = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < n - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int t = nums[i] + nums[i + 1] + nums[i + 2];
            if (t > target) {
                // 无论后面怎么选，选出的三个数的和不会比当前小
                if (t - target < Math.abs(ans - target)) {
                    ans = t;
                }
                break;
            }

            t = nums[i] + nums[n - 2] + nums[n - 1];
            if (t < target) {
                // 由于已经排序，当前和仍小于target，则无法找到比当前更优的答案，跳过后续步骤
                if (target - t < Math.abs(ans - target)) {
                    ans = t;
                }
                continue;
            }

            int j = i + 1, k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == target) return sum;
                if (Math.abs(sum - target) < Math.abs(ans - target)) ans = sum;
                if (sum < target) ++j;
                else --k;
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n^2)$
  • 空间复杂度：$O(logn)$