java - 一次迭代 4 个 HashMap 的键、值

Question

我有一个 HashMap 。

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

我想返回前 4 个键，然后是前 4 个值，然后重复直到什么都没有。

如何最好地做到这一点？

我整个上午都在尝试 :)

Answer 1

如果您可以设法将所有键作为列表获取，那么您可以一次迭代它们 4 个，因为这样您就可以通过索引位置获取它们，就像这样

//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);
}

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编辑

如果元素个数不是4的倍数，那么你可以这样做:

//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);
        }
    }
}