2 Sum

原題網址

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are NOT zero-based.

public int[] twoSum(int[] numbers, int target) {
// naive method O(n^2)
    // if ( numbers.length == 0 || numbers == null ) {
    //     return null;
    // }

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

    // return null;

    // HashMap O(n)

    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

    // for ( int i = 0 ; i < numbers.length ; i++ ) {
    //     if ( map.containsKey(target-numbers[i]) ) {
    //         return new int[]{map.get(target-numbers[i]), i+1};
    //     } else {
    //         map.put(numbers[i], i+1);
    //     }
    // }

    // return null;

    // Two pointer   ** For sorted array
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for ( int i = 0 ; i < numbers.length ; i++ ) {
        map.put(numbers[i], i+1);
    }
    Arrays.sort(numbers);
    int start = 0;
    int end = numbers.length-1;

    while ( start < end ) {
        if ( numbers[start] + numbers[end] == target ) {
            break;
        } else if ( numbers[start] + numbers[end] > target ) {
            end--;
        } else if ( numbers[start] + numbers[end] < target ) {
            start++;
        }
    }

    if ( start < end ) {
        int[] ans = new int[]{map.get(numbers[start]), map.get(numbers[end])};
        Arrays.sort(ans);
        return ans;
    } else {
        return null;
    }
}