【每日一题】- 2019-08-26 - 1122. 数组的相对排序

[azl397985856]

给你两个数组，arr1 和 arr2，

arr2 中的元素各不相同
arr2 中的每个元素都出现在 arr1 中
对 arr1 中的元素进行排序，使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾。

 

示例：

输入：arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
输出：[2,2,2,1,4,3,3,9,6,7,19]
 

提示：

arr1.length, arr2.length <= 1000
0 <= arr1[i], arr2[i] <= 1000
arr2 中的元素 arr2[i] 各不相同
arr2 中的每个元素 arr2[i] 都出现在 arr1 中

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/relative-sort-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

[asbstty]

resolve(arr1, arr2) {
    let curIndex = 0;
    for (const num of arr2) {
      for (let i = curIndex; i < arr1.length; i++) {
        const curNum = arr1[i];
        if (curNum === num) {
          const temp = arr1[i];
          arr1[i] = arr1[curIndex];
          arr1[curIndex] = temp;
          curIndex += 1;
        }
      }
    }
    for (let i = arr1.length - 1; i > curIndex; i--) {
      let flag = false;
      for (let j = curIndex; j < i; j++) {
        if (arr1[j] > arr1[j + 1]) {
          const temp = arr1[j];
          arr1[j] = arr1[j + 1];
          arr1[j + 1] = temp;
          flag = true;
        }
      }
      if (!flag) {
        break;
      }
    }
    return arr1;
  }

[raof01]

static vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
    auto m = unordered_map<int, int>();
    for (auto i = 0; i < arr2.size(); ++i) {
        m[arr2[i]] = i;
    }
    for (auto v : arr1) {
        if (m.find(v) == m.end()) {
            m[v] = 1000 + v;
        }
    }
    sort(arr1.begin(), arr1.end(), [&m](int a, int b) {
        return m[a] < m[b];
    });
    return arr1;
}

[wjj31767]

python3

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        d2 = {}
        for num,i in enumerate(arr2):
            d2[i] = num
        # print(d2)
        d1 = {}
        rest = []
        for i in arr1:
            if i in d2:
                if d2[i] not in d1:
                    d1[d2[i]] = [i,0]
                d1[d2[i]][1] += 1
            else:
                rest.append(i)
        # print(d1)
        res = []       
        for i in sorted(d1):
            for _ in range(d1[i][1]):
                res.append(d1[i][0])
        return res + sorted(rest)

[maninbule]

【c++】【STL】
一道简单题，我竟然做了好久555555

class Solution {
public:
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        map<int,int> mp;
        unordered_map<int,bool> st;
        vector<int> res;
        for(int i = 0;i<arr1.size();i++) mp[arr1[i]]++;
        for(int i = 0;i<arr2.size();i++) st[arr2[i]] = true;
        for(int i = 0;i<arr2.size();i++){
            while(mp[arr2[i]]--){
                res.push_back(arr2[i]);
            }
            
        }
        for(auto &m:mp){
            if(!st[m.first]){
                while(mp[m.first]--){
                    res.push_back(m.first);
                }
            }
        }
        return res;
    }
};

[maninbule]

看了上面这位兄弟的思路，感觉思路非常好,在此贴一下自己的cpp版本代码

class Solution {
public:
    unordered_map<int,int> mp;
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        for(int i = 0;i<arr2.size();i++) mp[arr2[i]] = i+1;
        for(int i = 0;i<arr1.size();i++) if(!mp[arr1[i]]) mp[arr1[i]] = arr1[i]+2000;
        sort(arr1.begin(),arr1.end(),[&](int &a,int &b){return mp[a]<mp[b];});
        return arr1;
    }
};

[stale]

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