多重 if 判断怎么优化？排序规则：等级从高到低，相同等级比较拥有同等级的数量，所有等级数量相等最后以得分进行比较

正确使用 likely 和 unlikely 能够明显提升分支预测的准确率。

if (map2.get("unitLevel") != null && map1.get("unitLevel") != null) {
int flag = compareLevel(map2.get("unitLevel").toString(), map1.get("unitLevel").toString());
if (flag != 0) {
return flag;
}
int flag2 = compareLevelCount((Integer) map2.get("countAA"), (Integer) map1.get("countAA"));
if (flag2 != 0) {
return flag2;
}
int flag3 = compareLevelCount((Integer) map2.get("countA"), (Integer) map1.get("countA"));
if (flag3 != 0) {
return flag3;
}
int flag4 = compareLevelCount((Integer) map2.get("countB"), (Integer) map1.get("countB"));
if (flag4 != 0) {
return flag4;
}

int flag5 = compareLevelCount((Integer) map2.get("countC"), (Integer) map1.get("countC"));
if (flag5 != 0) {
return flag5;
}
int flag6 = compareLevelCount((Integer) map2.get("countD"), (Integer) map1.get("countD"));
if (flag6 != 0) {
return flag6;
}
return Double.valueOf((Double) map2.get("unitScore")).compareTo(Double.valueOf((Double) map1
.get("unitScore")));
}

if(xxx != null){
//业务代码
}
可以改为
if(xxx == null){
return;
}
//业务代码

能跳出 if 包裹最好

map.get("xxx")是否有规律，如果有的话 你的 compareLevelCount 可以使用递归优化，显得代码整洁。但要写好注释并保证没有死循环
if (flag != 0) {
return flag;
}else{
return compareLevelCount((Integer) map2.get("xxx"), (Integer) map1.get("xxx"));
}

如上面朋友所说，看看 map 的 key 是否能遵从某种逻辑进行变换。可以的话，大体如下
public {returnType} {yourMethodName} ({yourParameters}) {
if (map2.get("unitLevel") == null || map1.get("unitLevel") == null) {
// return something according to your business logic
} else {
recursiveCompareLevelCount("first key0", "first key1", map1, map2);
}
}

private {returnType} recursiveCompareLevelCount(String key0, String key1, Map<K, V> map0, Map<K, V> map1) {
// key0 and key1 equals your end compare logic
if (key0.equals("Something") && key1.equals("Something")) {
return Double.valueOf((Double) map1.get(key0)).compareTo(Double.valueOf((Double) map0.get(key1)));
}

int flag = compareLevelCount((Integer) map1.get(key1), (Integer) map0.get(key0));
if (flag != 0) {
return flag;
} else {
// set some rule to increment the key
return compareLevelCount(new String("increment key0"), new String("increment key1"), map0, map1);
}
}

@xdchl 补充，主方法 else 少了个 return

@xdchl 最后 else 里面是调用 recursiveCompareLevelCount ，打错了

@xdchl 写完算了
import java.util.HashMap;
import java.util.Map;

public class MutipleIfQuestionFromV2ex {
public int testMethod() {
Map<String, Integer> map1 = new HashMap<>();
Map<String, Integer> map2 = new HashMap<>();
if (map2.get("unitLevel") == null || map1.get("unitLevel") == null) {
// return something according to your business logic
return 0;
} else {
return recursiveCompareLevelCount("countAA", "countAA", map1, map2);
}
}

private int recursiveCompareLevelCount(String key0, String key1, Map<String, Integer> map0, Map<String, Integer> map1) {
// key0 and key1 equals your end compare logic
if (key0.equals("unitScore") && key1.equals("unitScore")) {
return map1.get(key0).compareTo(map0.get(key1));
}

int flag = compareLevelCount(map1.get(key1), map0.get(key0));
if (flag != 0) {
return flag;
} else {
// set some rule to increment the key
char count = (char) (key0.charAt(key0.length() - 1) + 1);
if (count == 'E') {
key0 = "unitScore";
key1 = "unitScore";
} else if (count == 'B' && key0.charAt(key0.length() - 2) == 'A') {
key0 = "countA";
key1 = "countA";
} else {
key0 = key0.substring(0, key0.length() - 1) + count;
key1 = key1.substring(0, key1.length() - 1) + count;
}
return recursiveCompareLevelCount(key0, key1 , map0, map1);
}
}

private int compareLevelCount(int v1, int v2) {
return Integer.compare(v1, v2);
}
}