Exemple #1
0
        /*
         * source code from blog:
         * http://www.acmerblog.com/leetcode-two-sum-5223.html
         *
         * 解法1:先排序,然后从开头和结尾同时向中间查找,原理也比较简单。复杂度O(nlogn)
         *
         * julia's comment:
         *
         */
        public static ArrayList twoSum(int[] numbers, int target)
        {
            int len = numbers.Length;
            Node[] tmpNodes = new Node[len];

            for (int i = 0; i < len; i++)
            {
                tmpNodes[i].id = i + 1;
                tmpNodes[i].val = numbers[i];
            }

            Node[] nodes = sortArray(tmpNodes);

            int start = 0, end = len - 1;

            ArrayList ans = new ArrayList();

            while (start < end)
            {
                if (nodes[start].val + nodes[end].val == target)
                {

                    if (nodes[start].id > nodes[end].id)
                    {
                        swap(ref tmpNodes, start, end);

                        ans.Add(tmpNodes[start].id);
                        ans.Add(tmpNodes[end].id);

                        return ans;

                    }
                    else if (tmpNodes[start].val + tmpNodes[end].val < target)
                        start++;
                    else
                        end--;
                }
            }

            return ans;
        }
Exemple #2
0
 public static Node[] sortArray(Node[] nA)
 {
     return nA;   // implement later
 }
Exemple #3
0
 public static void swap(ref Node[] nA, int start, int end)
 {
     Node tmp = nA[start];
     nA[start] = nA[end];
     nA[end] = tmp;
 }
Exemple #4
0
 public static bool compare(Node a, Node  b)
 {
     return a.val < b.val;
 }