feat: solve No.1572

BaffinLee · BaffinLee · commit cd592312ef41 · 2023-05-08T23:41:27.000+08:00
diff --git a/1501-1600/1572. Matrix Diagonal Sum.md b/1501-1600/1572. Matrix Diagonal Sum.md
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+# 1572. Matrix Diagonal Sum
+
+- Difficulty: Easy.
+- Related Topics: Array, Matrix.
+- Similar Questions: Check if Every Row and Column Contains All Numbers, Check if Matrix Is X-Matrix.
+
+## Problem
+
+Given a square matrix `mat`, return the sum of the matrix diagonals.
+
+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.
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2020/08/14/sample_1911.png)
+
+```
+Input: mat = [[1,2,3],
+              [4,5,6],
+              [7,8,9]]
+Output: 25
+Explanation: Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
+Notice that element mat[1][1] = 5 is counted only once.
+```
+
+Example 2:
+
+```
+Input: mat = [[1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1],
+              [1,1,1,1]]
+Output: 8
+```
+
+Example 3:
+
+```
+Input: mat = [[5]]
+Output: 5
+```
+
+ 
+**Constraints:**
+
+
+	
+- `n == mat.length == mat[i].length`
+	
+- `1 <= n <= 100`
+	
+- `1 <= mat[i][j] <= 100`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[][]} mat
+ * @return {number}
+ */
+var diagonalSum = function(mat) {
+    var res = 0;
+    for (var i = 0; i < mat.length; i++) {
+        res += mat[i][i] + mat[i][mat.length - 1 - i];
+    }
+    if (mat.length % 2) {
+        var mid = Math.floor(mat.length / 2);
+        res -= mat[mid][mid];
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).
