LeetCode 219. Contains Duplicate II(C++版)
2017-03-31 13:39
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Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j]and the absolute difference
between i and j is at most k.
思路分析:
也就是说,有两个要求,i - j <= k && nums[i] == nums[j]
遍历数组,查找i和i+k是否相等
时间复杂度:O(n^2)
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
if(k <= 0) return false;
if(nums.size() == 0 || nums.size() == 1 ) return false;
for(int i = 0; i < nums.size(); i ++){
for(int j = i + 1; j - i <= k && j < nums.size(); j ++ )
if(nums[i] == nums[j]) return true;
}
return false;
}
};
结果超时。
方法二:
遍历数组,如果nums[ i ]已经在map中,判断其value和i的距离是否小于等于k,是,则返回true。否,则将nums[i]对应的value修改为当前的 i 值。
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
if(k <= 0) return false;
if(nums.size() == 0 || nums.size() == 1) return false;
//如果nums[ i ]已经在map中,判断其value和i的距离是否小于等于k,是,则返回true。否,则将nums[i]对应的value修改为当前的i值
unordered_map<int, int> cpnums;
for(int i = 0; i < nums.size(); i ++){
if(cpnums.find(nums[i]) == cpnums.end())
cpnums.insert({nums[i], i});
else{
if(i - cpnums[ nums[i]] <= k)
return true;
else
cpnums[nums[i]] = i;
}
}
return false;
}
};
通过。
between i and j is at most k.
思路分析:
也就是说,有两个要求,i - j <= k && nums[i] == nums[j]
遍历数组,查找i和i+k是否相等
时间复杂度:O(n^2)
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
if(k <= 0) return false;
if(nums.size() == 0 || nums.size() == 1 ) return false;
for(int i = 0; i < nums.size(); i ++){
for(int j = i + 1; j - i <= k && j < nums.size(); j ++ )
if(nums[i] == nums[j]) return true;
}
return false;
}
};
结果超时。
方法二:
遍历数组,如果nums[ i ]已经在map中,判断其value和i的距离是否小于等于k,是,则返回true。否,则将nums[i]对应的value修改为当前的 i 值。
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
if(k <= 0) return false;
if(nums.size() == 0 || nums.size() == 1) return false;
//如果nums[ i ]已经在map中,判断其value和i的距离是否小于等于k,是,则返回true。否,则将nums[i]对应的value修改为当前的i值
unordered_map<int, int> cpnums;
for(int i = 0; i < nums.size(); i ++){
if(cpnums.find(nums[i]) == cpnums.end())
cpnums.insert({nums[i], i});
else{
if(i - cpnums[ nums[i]] <= k)
return true;
else
cpnums[nums[i]] = i;
}
}
return false;
}
};
通过。
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