LeetCode-Shortest Word Distance II

因爲需要多次call 所以就存成hashmap 存position list

關鍵是找min的時候 實際上就是給兩個list 找其中數字最小difference 

方法就是每次更新那個小的 讓他變大 去靠近大的 （ 因爲list是sorted 從小到大）移動數字比較小的那個的指針

public class WordDistance {
    HashMap<String, List<Integer>> map = new HashMap<String, List<Integer>>();
    public WordDistance(String[] words) {
        for ( int i = 0; i < words.length; i ++ ){
            if (map.containsKey( words [ i ] )){
                map.get ( words[i] ).add ( i );
            }
            else{
                List<Integer> list = new ArrayList<Integer>();
                list.add ( i );
                map.put ( words[i], list );
            }
        }
    }

    public int shortest(String word1, String word2) {
        int min = Integer.MAX_VALUE;
        int p1 = 0;
        int p2 = 0;
        List<Integer> list1 = map.get ( word1 );
        List<Integer> list2 = map.get ( word2 );
        while ( p1 < list1.size() && p2 < list2.size() ){
            min = Math.min( min, Math.abs ( list1.get ( p1 ) - list2.get ( p2 ) ));
            if ( list1.get ( p1 ) > list2.get ( p2 ) )
                p2 ++;
            else 
                p1 ++;
        }
        return min;
    }
}