版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/u013132035/article/details/82154268
题目:
在一个长度为n的数组里的所有数字都在0到n-1的范围内。数组中某些数字是重复的,但不知道有几个数字重复了,也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。例如,如果输入长度为7的数组{2, 3, 1, 0, 2, 5, 3},那么对应的输出是重复的数字2或者3。
代码实现:
public static boolean getRepeatNums(int numbers[], int length, int duplication[]){
if(numbers == null || length <= 0){
return false;
}
for(int i = 0; i < length; i++){
if(numbers[i] < 0 || numbers[i] > length -1){
return false;
}
}
for(int i = 0; i < length; i++){
while(numbers[i] != i){
if(numbers[i] == numbers[numbers[i]]){
duplication[0] = numbers[i];
return true;
}
int tmp = numbers[i];
numbers[i] = numbers[tmp];
numbers[tmp] = tmp;
}
}
return false;
}
代码实现:
public static boolean getRepeatNums1(int numbers[], int length, int duplication[]){
if(numbers==null || length<=0){
return false;
}
for(int i = 0; i < length; i++){
if(numbers[i] < 0 || numbers[i] > length-1){
return false;
}
}
Set<Integer> set = new HashSet<Integer>();
for(int i = 0; i < length; i++){
if(set.contains(numbers[i])){
duplication[0] = numbers[i];
return true;
}else{
set.add(numbers[i]);
}
}
return false;
}
代码实现:
public boolean getRepeatNums2(int numbers[],int length,int [] duplication) {
HashMap<Integer, Integer> countMap = new HashMap<Integer, Integer>();
if(length < 2||numbers==null){
return false;
}
int j = 1;
for(int i = 0;i < length;i++){
if(countMap.get(numbers[i]) == null){
j = 1;
countMap.put(numbers[i], j);
}else{
j = countMap.get(numbers[i]);
j++;
countMap.put(numbers[i], j);
}
}
Iterator iter = countMap.entrySet().iterator();
while(iter.hasNext()){
Entry<Integer, Integer> entry = (Entry<Integer, Integer>) iter.next();
Integer key = entry.getKey();
Integer val = countMap.get(key);
if(val > 1){
duplication[0] = key;
return true;
}
}
return false;
}