From 4150454358d7b8601b369ba8ce335ee105b499a3 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Sat, 9 Nov 2019 11:21:11 +0800
Subject: [PATCH] Create 304. Range Sum Query 2D - Immutable.md

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+# [304. Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)
+# 思路
+题目要求一个二维区域和的检索，这其实是上一题[303. Range Sum Query - Immutable](https://github.com/ShusenTang/LeetCode/blob/master/solutions/303.%20Range%20Sum%20Query%20-%20Immutable.md)的二维版本，
+思路其实也是一样的。
+
+我们建立一个大小和nums一样的区域和数组sum，然后根据边界值的加减法来快速求出给定区域之和。其中sum[i][j]表示累计区间(0, 0)到(i, j)这个矩形区间所有的数字之和，
+那么此时如果我们想要快速求出(r1, c1)到(r2, c2)的矩形区间时，只需`sum[r2][c2] - sum[r2][c1 - 1] - sum[r1 - 1][c2] + sum[r1 - 1][c1 - 1]`即可。
+为了避免判断边界值，可以使用一个小技巧，把sum开辟大一点，并令sum[i+1][j+1]表示累计区间(0, 0)到(i, j)这个矩形区间所有的数字之和，见代码。
+
+# C++
+``` C++
+class NumMatrix {
+private:
+    int m, n;
+    vector<vector<int>>sum;
+    
+public:
+    NumMatrix(vector<vector<int>>& matrix) {
+        m = matrix.size();
+        n = m ? matrix[0].size() : 0;
+        if(!n) return;
+        sum = vector<vector<int>>(m + 1, vector<int>(n + 1, 0));
+        for(int i = 0; i < m; i++){
+            int row_sum = 0;
+            for(int j = 0; j < n; j++){
+                row_sum += matrix[i][j];
+                sum[i+1][j+1] = row_sum + sum[i][j+1];
+            }
+        }
+    }
+    
+    int sumRegion(int row1, int col1, int row2, int col2) {
+        return sum[row2+1][col2+1] -  sum[row2+1][col1] - sum[row1][col2+1] + sum[row1][col1];
+    }
+};
+```