LeetCode | Contains Duplicate II
2015-06-28 09:53
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Given an array of integers and an integer k,
find out whether there there are two distinct indices i and j in
the array such that nums[i] = nums[j] and
the difference between iand j is
at most k.
//要求判断数组是否有重复,且重复的两个元素的index的距离小于等于k
//直接的思路是用map<nums[i], index>来记录数组值及对应的index
public class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
boolean result = false;
Map<Integer, Integer> myMap = new HashMap<Integer, Integer>(nums.length);
for(int i=0; i<nums.length; i++){
if(myMap.containsKey(nums[i]) && i-myMap.get(nums[i]) <= k){
result = true;
break;
}
myMap.put(nums[i], i); //无论已遍历的集合是否containsKey,当不contains时为插入<nums[i],index>
} //当contains时,为更新已有<nums[i],index>映射中的index值
//即,当后put的key在map中已经存在时,就覆盖掉之前的<k,v>对
return result;
}
}
find out whether there there are two distinct indices i and j in
the array such that nums[i] = nums[j] and
the difference between iand j is
at most k.
//要求判断数组是否有重复,且重复的两个元素的index的距离小于等于k
//直接的思路是用map<nums[i], index>来记录数组值及对应的index
public class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
boolean result = false;
Map<Integer, Integer> myMap = new HashMap<Integer, Integer>(nums.length);
for(int i=0; i<nums.length; i++){
if(myMap.containsKey(nums[i]) && i-myMap.get(nums[i]) <= k){
result = true;
break;
}
myMap.put(nums[i], i); //无论已遍历的集合是否containsKey,当不contains时为插入<nums[i],index>
} //当contains时,为更新已有<nums[i],index>映射中的index值
//即,当后put的key在map中已经存在时,就覆盖掉之前的<k,v>对
return result;
}
}
//用长度k的窗口来表述间距k的限定条件,而不是用map来记录index public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { boolean result = false; Set<Integer> mySet = new HashSet<Integer>(nums.length); for(int i=0; i<nums.length; i++){ if(mySet.contains(nums[i])){ result = true; break; } mySet.add(nums[i]); if(i>=k){ //始终让set的大小为k,超了就将前面的元素删除 mySet.remove(nums[i-k]); } } return result; } }
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