# 2337. Move Pieces to Obtain a String
[2337. Move Pieces to Obtain a String](https://leetcode.com/problems/move-pieces-to-obtain-a-string/) (<font color="#FFB800"> Medium</font> 通過率: 55.6%)

## 限制條件
<ul>
<li>n == start.length == target.length</li>
<li>1 <= n <= 10^5</li>
<li>start and target consist of the characters 'L', 'R', and '_'.</li>
</ul>

### 解法 1
我的解法是先把兩個字串先記錄起來，並且記錄位置，接著再依照兩字串有沒有不一樣先篩選，再來就是如果L和R的位置不對，也步行的方式作為結果。

不過看起來解出來的複雜度很高而且記憶體的用量很大...

- 時間複雜度: O(n)
- 空間複雜度: O(n)

class Solution {
public:
    bool canChange(string start, string target) {
        string startRecord, targetRecord;
        vector<int> startPosition, targetPosition;
        for (int i = 0; i < start.size(); i++) {
            if (start[i] != '_') {
                startRecord.push_back(start[i]);
                startPosition.push_back(i);
            }
        }

        for (int i = 0; i < target.size(); i++) {
            if (target[i] != '_') {
                targetRecord.push_back(target[i]);
                targetPosition.push_back(i);
            }
        }

        if (startRecord != targetRecord)
            return false;

        for (int i = 0; i < targetRecord.size(); i++) {
            if ((targetRecord[i] == 'L') && (startPosition[i] < targetPosition[i]))
                return false;
            if ((targetRecord[i] == 'R') && (startPosition[i] > targetPosition[i]))
                return false;
        }
        return true;
    }
};

</int>