LeetCode-Shortest Word Distance II

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因为需要多次call 所以就存成hashmap 存position list

关键是找min的时候实际上就是给两个list 找其中数字最小difference

方法就是每次更新那个小的让他变大去靠近大的（因为list是sorted 从小到大）移动数字比较小的那个的指针

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;    }}

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