Question

Solution

题目大意:有一串隐藏的号码,另一个人会猜一串号码(数目相同),如果号码数字与位置都对了,给一个bull,数字对但位置不对给一个cow,注:数字对与位置对优先,一个号码不能重复判断.

思路:构造map结构,遍历实现

Java实现:实现的不漂亮,好歹能通过

public String getHint(String secret, String guess) {    Map
    
      map = new HashMap<>();    for (int i=0; i
     
       cowsList = new ArrayList<>(); // for count cows    // count bulls    for (int i=0; i
      
        idxList;    int count;    // constructor    public Index() {        idxList = new ArrayList<>();        count = 0;    }    void add(int x) {        idxList.add(x);        count++;    }    boolean isCow() {        return count-- > 0;    }    boolean isBull(int x) {        for (int tmp : idxList) {            if (x == tmp) {                count--;                return true;            }        }        return false;    }}
      
     
    

Ref

public String getHint(String secret, String guess) {    int bulls = 0;    int cows = 0;    int[] numbers = new int[10];    for (int i = 0; i
    
      0) cows++;            numbers[s] ++;            numbers[g] --;        }    }    return bulls + "A" + cows + "B";}
    

public String getHint(String secret, String guess) {    int bulls = 0;    int cows = 0;    int[] numbers = new int[10];    for (int i = 0; i
    
      0) cows++;        }    }    return bulls + "A" + cows + "B";}