一刷861

Author: diguage

README.adoc

@@ -6051,14 +6051,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0860-lemonade-change.adoc[题解]
 //|Easy
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/score-after-flipping-matrix/[861. Score After Flipping Matrix^]
-//|{source_base_url}/_0861_ScoreAfterFlippingMatrix.java[Java]
-//|{doc_base_url}/0861-score-after-flipping-matrix.adoc[题解]
-//|Medium
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/score-after-flipping-matrix/[861. Score After Flipping Matrix^]
+|{source_base_url}/_0861_ScoreAfterFlippingMatrix.java[Java]
+|{doc_base_url}/0861-score-after-flipping-matrix.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/shortest-subarray-with-sum-at-least-k/[862. Shortest Subarray with Sum at Least K^]
 //|{source_base_url}/_0862_ShortestSubarrayWithSumAtLeastK.java[Java]

docs/0861-score-after-flipping-matrix.adoc

@@ -1,49 +1,70 @@
 [#0861-score-after-flipping-matrix]
-= 861. Score After Flipping Matrix
+= 861. 翻转矩阵后的得分
 
-{leetcode}/problems/score-after-flipping-matrix/[LeetCode - Score After Flipping Matrix^]
+https://leetcode.cn/problems/score-after-flipping-matrix/[LeetCode - 861. 翻转矩阵后的得分^]
 
-We have a two dimensional matrix `A` where each value is `0` or `1`.
+给你一个大小为 `m x n` 的二元矩阵 `grid` ，矩阵中每个元素的值为 `0` 或 `1`。
 
-A move consists of choosing any row or column, and toggling each value in that row or column: changing all `0`s to `1`s, and all `1`s to `0`s.
+一次 *移动* 是指选择任一行或列，并转换该行或列中的每一个值：将所有 `0` 都更改为 `1`，将所有 `1` 都更改为 `0`。
 
-After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.
+在做出任意次数的移动后，将该矩阵的每一行都按照二进制数来解释，矩阵的 *得分* 就是这些数字的总和。
 
-Return the highest possible score.
+在执行任意次 *移动* 后（含 0 次），返回可能的最高分数。
 
- 
+*示例 1：*
 
+image::images/0861-01.jpg[{image_attr}]
 
+....
+输入：grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
+输出：39
+解释：0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
+....
 
+*示例 2：*
 
+....
+输入：grid = [[0]]
+输出：1
+....
 
-*Example 1:*
-
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
-*Output:* 39
-*Explanation:*
-Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
-0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
-----
-
- 
-
-*Note:*
+*提示：*
 
-
-. `1 <= A.length <= 20`
-. `1 <= A[0].length <= 20`
-. `A[i][j]` is `0` or `1`.
+* `m == grid.length`
+* `n == grid[i].length`
+* `1 \<= m, n \<= 20+`
+* `grid[i][j]` 为 `0` 或 `1`
 
 
 
+== 思路分析
 
 
 [[src-0861]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0861_ScoreAfterFlippingMatrix.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0861_ScoreAfterFlippingMatrix_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/score-after-flipping-matrix/solutions/512802/jiu-ying-suan-tong-su-yi-dong-mei-xiang-fgt9j/[861. 翻转矩阵后的得分 - 就硬算，通俗易懂，一开始没想到位操作^]
+. https://leetcode.cn/problems/score-after-flipping-matrix/solutions/512323/fan-zhuan-ju-zhen-tan-xin-xin-lu-li-chen-21h7/[861. 翻转矩阵后的得分 - 翻转矩阵贪心心路历程！^]
+. https://leetcode.cn/problems/score-after-flipping-matrix/solutions/511825/fan-zhuan-ju-zhen-hou-de-de-fen-by-leetc-cxma/[861. 翻转矩阵后的得分 - 官方题解^]

docs/images/0861-01.jpg

@@ -1814,7 +1814,7 @@ include::0859-buddy-strings.adoc[leveloffset=+1]
 
 // include::0860-lemonade-change.adoc[leveloffset=+1]
 
-// include::0861-score-after-flipping-matrix.adoc[leveloffset=+1]
+include::0861-score-after-flipping-matrix.adoc[leveloffset=+1]
 
 // include::0862-shortest-subarray-with-sum-at-least-k.adoc[leveloffset=+1]
 

docs/index.adoc

@@ -1152,6 +1152,11 @@ endif::[]
 |{doc_base_url}/0859-buddy-strings.adoc[题解]
 |✅ 哈希。注意处理字符串相同的情况。
 
+|{counter:codes}
+|{leetcode_base_url}/score-after-flipping-matrix/[861. Score After Flipping Matrix^]
+|{doc_base_url}/0861-score-after-flipping-matrix.adoc[题解]
+|✅ 贪心。确保最高位为 `1`，确保每一列的 `1` 占多数。
+
 
 |===
 

logbook/202601.adoc

@@ -0,0 +1,52 @@
+package com.diguage.algo.leetcode;
+
+public class _0861_ScoreAfterFlippingMatrix {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2026-08-12 22:30:19
+   */
+  public int matrixScore(int[][] grid) {
+    int m = grid.length;
+    int n = grid[0].length;
+    // 确保第一行都是 1
+    for (int[] row : grid) {
+      if (row[0] == 1) {
+        continue;
+      }
+      for (int i = 0; i < n; i++) {
+        row[i] ^= 1;
+      }
+    }
+    // 确保每一列的 1 占多数
+    for (int i = 0; i < n; i++) {
+      int cnt = 0;
+      for (int[] ints : grid) {
+        cnt += ints[i];
+      }
+      if (m / 2 < cnt) {
+        continue;
+      }
+      for (int j = 0; j < m; j++) {
+        grid[j][i] ^= 1;
+      }
+    }
+    // 转成数字。⚠️：这一步可以和上一步合并，减少一次循环
+    int result = 0;
+    for (int[] row : grid) {
+      int cnt = 0;
+      for (int i = 0; i < n; i++) {
+        cnt = cnt * 2 + row[i];
+      }
+      result += cnt;
+    }
+    return result;
+  }
+  // end::answer[]
+
+  static void main() {
+    new _0861_ScoreAfterFlippingMatrix()
+//      .matrixScore(new int[][]{{0, 0, 1, 1}, {1, 0, 1, 0}, {1, 1, 0, 0}});
+      .matrixScore(new int[][]{{0, 1}, {1, 1}});
+  }
+}