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‎docs/AimForOffer/数据结构相关/2_字符串.md‎

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最朴素的方法就是依次从待匹配串的每一个位置开始，逐一与模版串匹配，
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因为最多检查 (n - m)个位置，所以方法的复杂度为 O(m*(n-1))。
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/string/str_1.png" width="500"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/string/str_1.png" width="500"/></div>
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```java
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//haystack 是待匹配串

‎docs/AimForOffer/算法思想相关/4_搜索.md‎

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[N皇后](https://leetcode-cn.com/problems/n-queens/)
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_22.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_22.png" width="600"/></div>
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_23.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_23.png" width="600"/></div>
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- 思路：
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_24.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_24.png" width="600"/></div>
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_25.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_25.png" width="600"/></div>
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在第 i 行放上皇后，判断是否其他地方能够放皇后的情况：
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对角线1总共有 (2*n-1) 条；m 与 i 、 j 的关系：m=i+j。
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_26.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_26.png" width="600"/></div>
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(3)对角线2：dai2[m]表示第 m 条对角线被2占用，如 dai2[0]=true 表示[0,3] 位置不能放皇后，
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dai2[2]=true 则表示 [0,2] 和 [1,3] 不能放皇后,
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对角线2总共有（2*n-1）条；m 与 i、j 关系：m=i-j+n-1。
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_27.png" width="600"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_27.png" width="600"/></div>
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```java
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private List<List<String>> res;

‎docs/AimForOffer/算法思想相关/5_排列组合.md‎

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[全排列](https://leetcode-cn.com/problems/permutations/)
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<div align="center"><img src="https://gitee.com/duhouan/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_5.png" width="500px"/></div>
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<div align="center"><img src="https://github.com/DuHouAn/ImagePro/raw/master/java-notes/leetcode/backtrack//backtrack_5.png" width="500px"/></div>
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```java
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private List<List<Integer>> res;

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