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LeetCode 题解 1572, 1573
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# [1572. 矩阵对角线元素的和](https://leetcode-cn.com/problems/matrix-diagonal-sum)
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[English Version](/solution/1500-1599/1572.Matrix Diagonal Sum/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个正方形矩阵 <code>mat</code>,请你返回矩阵对角线元素的和。</p>
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<p>请你返回在矩阵主对角线上的元素和副对角线上且不在主对角线上元素的和。</p>
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<p>&nbsp;</p>
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<p><strong>示例&nbsp; 1:</strong></p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png" style="height:174px; width:336px" /></p>
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<pre>
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<strong>输入:</strong>mat = [[<strong>1</strong>,2,<strong>3</strong>],
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&nbsp; [4,<strong>5</strong>,6],
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&nbsp; [<strong>7</strong>,8,<strong>9</strong>]]
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<strong>输出:</strong>25
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<strong>解释:</strong>对角线的和为:1 +たす 5 +たす 9 +たす 3 +たす 7 =わ 25
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请注意,元素 mat[1][1] = 5 只会被计算一次。
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</pre>
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<p><strong>示例&nbsp; 2:</strong></p>
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<pre>
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<strong>输入:</strong>mat = [[<strong>1</strong>,1,1,<strong>1</strong>],
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&nbsp; [1,<strong>1</strong>,<strong>1</strong>,1],
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&nbsp; [1,<strong>1</strong>,<strong>1</strong>,1],
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&nbsp; [<strong>1</strong>,1,1,<strong>1</strong>]]
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<strong>输出:</strong>8
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>mat = [[<strong>5</strong>]]
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<strong>输出:</strong>5
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>n == mat.length == mat[i].length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 100</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [1572. Matrix Diagonal Sum](https://leetcode.com/problems/matrix-diagonal-sum)
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[中文文档](/solution/1500-1599/1572.Matrix Diagonal Sum/README.md)
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## Description
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<p>Given a&nbsp;square&nbsp;matrix&nbsp;<code>mat</code>, return the sum of the matrix diagonals.</p>
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<p>Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png" style="width: 336px; height: 174px;" />
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<pre>
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<strong>Input:</strong> mat = [[<strong>1</strong>,2,<strong>3</strong>],
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&nbsp; [4,<strong>5</strong>,6],
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&nbsp; [<strong>7</strong>,8,<strong>9</strong>]]
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<strong>Output:</strong> 25
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<strong>Explanation: </strong>Diagonals sum: 1 +たす 5 +たす 9 +たす 3 +たす 7 =わ 25
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Notice that element mat[1][1] = 5 is counted only once.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> mat = [[<strong>1</strong>,1,1,<strong>1</strong>],
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&nbsp; [1,<strong>1</strong>,<strong>1</strong>,1],
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&nbsp; [1,<strong>1</strong>,<strong>1</strong>,1],
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&nbsp; [<strong>1</strong>,1,1,<strong>1</strong>]]
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<strong>Output:</strong> 8
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> mat = [[<strong>5</strong>]]
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<strong>Output:</strong> 5
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == mat.length == mat[i].length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 100</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public int diagonalSum(int[][] mat) {
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int sum = 0, n = mat.length, mid = n >> 1;
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for (int i = 0, j = n - 1; i < n; i++, j--) {
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sum += (mat[i][i] + mat[i][j]);
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}
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return n % 2 == 0 ? sum : sum - mat[mid][mid];
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}
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}
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# [1573. 分割字符串的方案数](https://leetcode-cn.com/problems/number-of-ways-to-split-a-string)
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[English Version](/solution/1500-1599/1573.Number of Ways to Split a String/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个二进制串&nbsp;<code>s</code>&nbsp; (一个只包含 0 和 1 的字符串),我们可以将 <code>s</code>&nbsp;分割成 3 个 <strong>非空</strong>&nbsp;字符串 s1, s2, s3 (s1 + s2 + s3 = s)。</p>
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<p>请你返回分割&nbsp;<code>s</code>&nbsp;的方案数,满足 s1,s2 和 s3 中字符 &#39;1&#39; 的数目相同。</p>
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<p>由于答案可能很大,请将它对 10^9 + 7 取余后返回。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>s = &quot;10101&quot;
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<strong>输出:</strong>4
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<strong>解释:</strong>总共有 4 种方法将 s 分割成含有 &#39;1&#39; 数目相同的三个子字符串。
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&quot;1|010|1&quot;
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&quot;1|01|01&quot;
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&quot;10|10|1&quot;
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&quot;10|1|01&quot;
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>s = &quot;1001&quot;
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<strong>输出:</strong>0
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong>s = &quot;0000&quot;
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<strong>输出:</strong>3
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<strong>解释:</strong>总共有 3 种分割 s 的方法。
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&quot;0|0|00&quot;
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&quot;0|00|0&quot;
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&quot;00|0|0&quot;
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre><strong>输入:</strong>s = &quot;100100010100110&quot;
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<strong>输出:</strong>12
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>s[i] == &#39;0&#39;</code>&nbsp;或者&nbsp;<code>s[i] == &#39;1&#39;</code></li>
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<li><code>3 &lt;= s.length &lt;= 10^5</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [1573. Number of Ways to Split a String](https://leetcode.com/problems/number-of-ways-to-split-a-string)
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[中文文档](/solution/1500-1599/1573.Number of Ways to Split a String/README.md)
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## Description
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<p>Given a binary string <code>s</code> (a string consisting only of &#39;0&#39;s and &#39;1&#39;s),&nbsp;we can split <code>s</code>&nbsp;into 3 <strong>non-empty</strong> strings s1, s2, s3 (s1+ s2+ s3 = s).</p>
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<p>Return the number of ways <code>s</code> can be split such that the number of&nbsp;characters &#39;1&#39; is the same in s1, s2, and s3.</p>
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<p>Since the answer&nbsp;may be too large,&nbsp;return it modulo&nbsp;10^9 + 7.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;10101&quot;
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> There are four ways to split s in 3 parts where each part contain the same number of letters &#39;1&#39;.
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&quot;1|010|1&quot;
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&quot;1|01|01&quot;
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&quot;10|10|1&quot;
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&quot;10|1|01&quot;
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;1001&quot;
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<strong>Output:</strong> 0
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;0000&quot;
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> There are three ways to split s in 3 parts.
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&quot;0|0|00&quot;
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&quot;0|00|0&quot;
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&quot;00|0|0&quot;
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;100100010100110&quot;
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<strong>Output:</strong> 12
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 &lt;= s.length &lt;= 10^5</code></li>
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<li><code>s[i]</code> is <code>&#39;0&#39;</code>&nbsp;or&nbsp;<code>&#39;1&#39;</code>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public int numWays(String s) {
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char[] chars = s.toCharArray();
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List<Long> p = new ArrayList<>();
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for (int i = 0; i < chars.length; i++) {
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if (chars[i] == '1') {
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p.add((long) i);
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}
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}
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int l = p.size();
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if (l % 3 != 0) {
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return 0;
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}
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int MOD = (int) (1e9 + 7);
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if (l == 0) {
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return (int) (((long) (s.length() - 1) * (s.length() - 2) / 2) % MOD);
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}
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// 每 n/3 的地方为分界线
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return (int) ((p.get(l / 3) - p.get(l / 3 - 1)) * (p.get(2 * l / 3) - p.get(2 * l / 3 - 1))
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% MOD);
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}
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}

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