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1331. Rank Transform of an Array

中文文档

Description

Given an array of integers arr, replace each element with its rank.

The rank represents how large the element is. The rank has the following rules:

  • Rank is an integer starting from 1.
  • The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
  • Rank should be as small as possible.

 

Example 1:

Input: arr = [40,10,20,30]
Output: [4,1,2,3]
Explanation: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.

Example 2:

Input: arr = [100,100,100]
Output: [1,1,1]
Explanation: Same elements share the same rank.

Example 3:

Input: arr = [37,12,28,9,100,56,80,5,12]
Output: [5,3,4,2,8,6,7,1,3]

 

Constraints:

  • 0 <= arr.length <= 105
  • -109 <= arr[i] <= 109

Solutions

Python3

class Solution:
    def arrayRankTransform(self, arr: List[int]) -> List[int]:
        t = sorted(set(arr))
        return [bisect_left(t, x) + 1 for x in arr]
class Solution:
    def arrayRankTransform(self, arr: List[int]) -> List[int]:
        m = {v: i for i, v in enumerate(sorted(set(arr)), 1)}
        return [m[v] for v in arr]

Java

class Solution {
    public int[] arrayRankTransform(int[] arr) {
        Set<Integer> s = new HashSet<>();
        for (int v : arr) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort((a, b) -> a - b);
        Map<Integer, Integer> m = new HashMap<>();
        for (int i = 0; i < alls.size(); ++i) {
            m.put(alls.get(i), i + 1);
        }
        int[] ans = new int[arr.length];
        for (int i = 0; i < arr.length; ++i) {
            ans[i] = m.get(arr[i]);
        }
        return ans;
    }
}
class Solution {
    public int[] arrayRankTransform(int[] arr) {
        Set<Integer> s = new HashSet<>();
        for (int v : arr) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort((a, b) -> a - b);
        int n = arr.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = Collections.binarySearch(alls, arr[i]) + 1;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> arrayRankTransform(vector<int>& arr) {
        unordered_set<int> s(arr.begin(), arr.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        unordered_map<int, int> m;
        for (int i = 0; i < alls.size(); ++i) m[alls[i]] = i + 1;
        vector<int> ans;
        for (int v : arr) ans.push_back(m[v]);
        return ans;
    }
};
class Solution {
public:
    vector<int> arrayRankTransform(vector<int>& arr) {
        unordered_set<int> s(arr.begin(), arr.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        vector<int> ans;
        for (int v: arr) ans.push_back(lower_bound(alls.begin(), alls.end(), v) - alls.begin() + 1);
        return ans;
    }
};

Go

func arrayRankTransform(arr []int) []int {
	s := make(map[int]bool)
	for _, v := range arr {
		s[v] = true
	}
	var alls []int
	for v := range s {
		alls = append(alls, v)
	}
	sort.Ints(alls)
	m := make(map[int]int)
	for i, v := range alls {
		m[v] = i + 1
	}
	var ans []int
	for _, v := range arr {
		ans = append(ans, m[v])
	}
	return ans
}

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