迭代鍵，一次HashMap 4的值

Question

我有一個hashmap。

public HashMap<String, Integer> lines = new HashMap<String, Integer>();

我想返回前4個鍵，然后是前4個值並重復直到沒有任何東西。

怎么做到最好？

我整個上午都在嘗試:)

Answer 1

如果您可以設法將所有鍵作為列表獲取，那么您可以一次迭代4個，因為這樣您就可以通過索引位置獲取它們，就像這樣

//get all keys as list
List<String> list = new ArrayList<String>(lines.keySet());

//one iteration of this loop deals with 4 keys and 4 values.
for(i=0; i<n; i=i+4) {
    k1 = list.get(i);
    v1 = lines.get(k1);

    k2 = list.get(i+1);
    v2 = lines.get(k2);

    k3 = list.get(i+2);
    v3 = lines.get(k3);

    k4 = list.get(i+3);
    v4 = lines.get(k4);
}

---

編輯

如果元素的數量不是4的倍數，那么你可以這樣做：

//get all keys as list
List<String> list = new ArrayList<String>(lines.keySet());

//one iteration of this loop deals with 4 keys and 4 values.
int mod = n%4;
for(i=0; i<n-mod; i=i+4) {
    k1 = list.get(i);
    v1 = lines.get(k1);

    k2 = list.get(i+1);
    v2 = lines.get(k2);

    k3 = list.get(i+2);
    v3 = lines.get(k3);

    k4 = list.get(i+3);
    v4 = lines.get(k4);
}

//deal with last 1, 2 or 3 elements separately.
if(mod>=1) {
    k1 = list.get(i);
    v1 = lines.get(k1);

    if(mod>=2) {
        k2 = list.get(i+1);
        v2 = lines.get(k2);

        if(mod>=3) {
            k3 = list.get(i+2);
            v3 = lines.get(k3);
        }
    }
}