LeetCode 1 Two Sum

原题如下：

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

最直接的方法是两层循环遍历数组，对数组中所有元素一一相加并比对结果，时间复杂度是O(n^2)。代码如下：

    public int[] twoSum1(final int[] numbers, int target) {
        for (int i = 0; i < numbers.length; i++) {
            for (int j = i+1; j < numbers.length; j++) {
                int sum = numbers[i] + numbers[j];
                if(sum == target){
                    return new int[]{i,j};
                }
            }
        }
        return null;
    }

优化之后时间复杂度可以降为O(nlogn)，具体方法是先对数组进行排序（nlogn），然后分别使用两个指针从数组头和数组尾开始查找（n），利用数组有序的特性可以在O(n)的时间复杂度内完成检索。代码如下：

    public int[] twoSum2(final int[] numbers, int target) {
        List<Integer> indexes = new ArrayList<Integer>();
        for(int i = 0; i < numbers.length ; i++){
            indexes.add(i);
        }
        Collections.sort(indexes, new Comparator<Integer>(){
            @Override
            public int compare(Integer o1, Integer o2) {
                return numbers[o1]-numbers[o2];
            }
        });
        int i=0,j=numbers.length-1;
        while(i < j){
            int sum = numbers[indexes.get(i)] + numbers[indexes.get(j)];
            if(sum == target){
                break;
            }else if(sum < target){
                i++;
            }else{
                j--;
            }
        }
        if(i < j){
            int i1 = indexes.get(i);
            int i2 = indexes.get(j);
            return i1< i2? new int[]{i1,i2}:new int[]{i2,i1};
        }
        return null;
    }

参考源码：https://github.com/pkufork/Martians/blob/master/src/main/java/com/pkufork/martians/leetcode/L1_TwoSum.java