1365. How Many Numbers Are Smaller Than the Current Number*
https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
题目描述
Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j’s such that j != i and nums[j] < nums[i].
Return the answer in an array.
Example 1:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:
Input: nums = [6,5,4,8]
Output: [2,1,0,3]
Example 3:
Input: nums = [7,7,7,7]
Output: [0,0,0,0]
Constraints:
-
2 <= nums.length <= 500 -
0 <= nums[i] <= 100
C++ 实现 1
使用哈希表, 如果已经访问过的数据, 那么直接从哈希表中取结果.
class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
vector<int> res;
unordered_map<int, int> record;
for (int i = 0; i < nums.size(); ++ i) {
int count = 0;
if (!record.count(nums[i])) {
for (int j = 0; j < nums.size(); ++ j)
if (j != i)
if (nums[j] < nums[i]) count ++;
record[nums[i]] = count;
}
res.push_back(record[nums[i]]);
}
return res;
}
};
C++ 实现 2
也可以使用排序+哈希表.
下面代码来自 LeetCode Submission.
class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
vector<int> originNums(nums);
sort(nums.begin(), nums.end());
unordered_map<int, int> temp;
for(int i = 0; i < nums.size() ; i++){
if(temp.find(nums[i] )== temp.end()){
temp[nums[i]] = i;
}
}
for(int i = 0; i < originNums.size();i++){
originNums[i] = temp[originNums[i]];
}
return originNums;
}
};
