[leetcode] 16. 最接近的三数之和

跟之前几个题思路大致相同，
贪心中的双指针法

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);

        int min = Integer.MAX_VALUE;
        int ans = 0;
        int flag = 0;

        for (int base = 0; base < nums.length; base++) {
            int i = base + 1, j = nums.length - 1;
            if (flag == 1) break;
            while (i < j) {
                int tmp = nums[base] + nums[i] + nums[j];
                if (Math.abs(tmp - target) < min) {
                    ans = tmp;
                    min = Math.abs(tmp - target);
                }
                if (tmp > target) {
                    j--;
                } else if (tmp < target) {
                    i++;
                } else {
                    flag = 1;
                    break;
                }
            }
        }

        return ans;

    }
}

更新：
观摩了下榜单第一老哥的代码，写的不错,比我的快了许多：

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int n = nums.length;
        Arrays.sort(nums);
        int less = nums[0] + nums[1] + nums[2];
        int more = nums[n-3] + nums[n-2] + nums[n-1];
        if (less >= target)
            return less;
        if (more <= target)
            return more;
        for (int i = 0; i < n - 2; i++){
            int min = nums[i] + nums[i+1] + nums[i+2];
            int max = nums[i] + nums[n-2] + nums[n-1];
            if (min > target){
                more = Math.min(more, min);
                break;
            }
            if (max < target){
                less = Math.max(less, max);
                continue;
            }
            int j = i + 1, k = n - 1;
            while(j < k){
                int sum = nums[i] + nums[j] + nums[k];
                if(sum == target) return target;
                else if(sum > target){
                    more = Math.min(more, sum);
                    while(--k > j && nums[k]==nums[k+1]);
                }else{
                    less = Math.max(less, sum);
                    while(++j < k && nums[j]==nums[j-1]);
                }
            }
        }
        return more - target > target - less ? less : more;
    }
}