1074:元素和为目标值的子矩阵数量

src/main/java/leetcode/editor/cn/NumberOfSubmatricesThatSumToTarget.java

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+//给出矩阵 matrix 和目标值 target，返回元素总和等于目标值的非空子矩阵的数量。 
+//
+// 子矩阵 x1, y1, x2, y2 是满足 x1 <= x <= x2 且 y1 <= y <= y2 的所有单元 matrix[x][y] 的集合。 
+//
+//
+// 如果 (x1, y1, x2, y2) 和 (x1', y1', x2', y2') 两个子矩阵中部分坐标不同（如：x1 != x1'），那么这两个子矩阵
+//也不同。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//
+// 
+//输入：matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
+//输出：4
+//解释：四个只含 0 的 1x1 子矩阵。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：matrix = [[1,-1],[-1,1]], target = 0
+//输出：5
+//解释：两个 1x2 子矩阵，加上两个 2x1 子矩阵，再加上一个 2x2 子矩阵。
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：matrix = [[904]], target = 0
+//输出：0
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= matrix.length <= 100 
+// 1 <= matrix[0].length <= 100 
+// -1000 <= matrix[i] <= 1000 
+// -10^8 <= target <= 10^8 
+// 
+// Related Topics 数组 动态规划 Sliding Window 
+// 👍 117 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+//1074:元素和为目标值的子矩阵数量
+class NumberOfSubmatricesThatSumToTarget {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new NumberOfSubmatricesThatSumToTarget().new Solution();
+        System.out.println(solution.numSubmatrixSumTarget(new int[][]{{0,1,0}, {1,1,1},{0,1,0}}, 0));
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int numSubmatrixSumTarget(int[][] matrix, int target) {
+            int xLength = matrix.length;
+            int yLength = matrix[0].length;
+            int[][] sum = new int[xLength + 1][yLength + 1];
+            for (int i = 1; i <= xLength; ++i) {
+                for (int j = 1; j <= yLength; ++j) {
+                    sum[i][j] = matrix[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
+                }
+            }
+            int count = 0;
+            for (int i = 1; i <= xLength; ++i) {
+                for (int j = 1; j <= yLength; ++j) {
+                    for (int k = 1; k <= i; ++k) {
+                        for (int l = 1; l <= j; ++l) {
+                            if (sum[i][j] - sum[k - 1][j] - sum[i][l - 1] + sum[k - 1][l - 1] == target){
+                                ++count;
+                            }
+                        }
+                    }
+                }
+            }
+            return count;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/NumberOfSubmatricesThatSumToTarget.md

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+<p>给出矩阵 <code>matrix</code> 和目标值 <code>target</code>，返回元素总和等于目标值的非空子矩阵的数量。</p>
+
+<p>子矩阵 <code>x1, y1, x2, y2</code> 是满足 <code>x1 <= x <= x2</code> 且 <code>y1 <= y <= y2</code> 的所有单元 <code>matrix[x][y]</code> 的集合。</p>
+
+<p>如果 <code>(x1, y1, x2, y2)</code> 和 <code>(x1', y1', x2', y2')</code> 两个子矩阵中部分坐标不同（如：<code>x1 != x1'</code>），那么这两个子矩阵也不同。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg" style="width: 242px; height: 242px;" /></p>
+
+<pre>
+<strong>输入：</strong>matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
+<strong>输出：</strong>4
+<strong>解释：</strong>四个只含 0 的 1x1 子矩阵。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>matrix = [[1,-1],[-1,1]], target = 0
+<strong>输出：</strong>5
+<strong>解释：</strong>两个 1x2 子矩阵，加上两个 2x1 子矩阵，再加上一个 2x2 子矩阵。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>matrix = [[904]], target = 0
+<strong>输出：</strong>0
+</pre>
+
+<p> </p>
+
+<p><strong><strong>提示：</strong></strong></p>
+
+<ul>
+	<li><code>1 <= matrix.length <= 100</code></li>
+	<li><code>1 <= matrix[0].length <= 100</code></li>
+	<li><code>-1000 <= matrix[i] <= 1000</code></li>
+	<li><code>-10^8 <= target <= 10^8</code></li>
+</ul>
+<div><div>Related Topics</div><div><li>数组</li><li>动态规划</li><li>Sliding Window</li></div></div>\n<div><li>👍 116</li><li>👎 0</li></div>

src/main/java/leetcode/editor/cn/NumberOfSubmatricesThatSumToTarget181859893.txt

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+    class Solution {
+        public int numSubmatrixSumTarget(int[][] matrix, int target) {
+            int xLength = matrix.length;
+            int yLength = matrix[0].length;
+            int[][] sum = new int[xLength + 1][yLength + 1];
+            Map<Integer, Integer> map = new HashMap<>();
+            int count = 0;
+            for (int i = 1; i < xLength + 1; i++) {
+                for (int j = 1; j < yLength + 1; j++) {
+                    sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
+                    if (matrix[i - 1][j - 1] == target) {
+                        count++;
+                    }
+                    map.put(sum[i][j], map.getOrDefault(sum[i][j], 0));
+                }
+            }
+            for (int key : map.keySet()) {
+                int value = map.get(key);
+                if (key == 2 * target && value > 1) {
+                    count += value * (value - 1) / 2;
+                } else if (key > target && map.containsKey(key - target)) {
+                    count += value * map.get(key - target);
+                }
+            }
+            return count;
+        }
+    }
+
+//total_testcases:40
+//total_correct:3
+//input_formatted:[[1,-1],[-1,1]]
+0
+//expected_output:5
+//code_output:0