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# 30. Substring with Concatenation of All Words

## 1.問題

* 給一個字串, 還有一個字的list, 每個字的長度相同. 找出所有字所拼成的子字串在字串中的起始索引

![](/files/-LNokMG20kdPnH2GMxaD)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
* function header, parameter
* test input
* 說明想法
  * 此題的目的是確認字串中是否有符合目標的子字串: 除了用字串相等外, 可以用範圍內的substr的字串是否有在map中出現, 而且次數相同, 如滿足此條件則將索引push到vector中
  * 用兩個map: 第一個存放string list, 第二個map存放每次移動時, substr中字串所出現的次數
* 測試計算複雜度: O(n)

## **3.程式碼** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> res;
        int size = words.size(); 
        if (size== 0 || s.empty() || size* words[0].length() > s.length()) {
            return res;
        }
        
        map<string, int> m1;
        for (int i = 0; i < size; i++) {
            m1[words[i]]++;
        }
        
        for (int i = 0 ; i <= s.length() - size * words[0].length(); i++) {
            map<string, int> m2;
            bool flag = true;
            for (int j = 0; j < size; j++) {
                string tmp = s.substr(i + j * words[0].length(), words[0].length());
                if (m1.find(tmp) != m1.end()) {
                    m2[tmp]++;
                } else {
                    flag = false;
                    break;
                }
                
                if (m2[tmp] > m1[tmp]) {
                    flag = false;
                    break;
                }
            }
            if (flag) {
                res.push_back(i);
            }
        }
        
        return res;
    }
};
```