前缀和

content/posts/algorithm/data structure/数组-前缀和数组.md

@@ -6,16 +6,23 @@ series: [data structure, trick]
 categories: [algorithm]
 ---
 
-## 基础
+## 概念
 
 应用场景：在原始数组不会被修改的情况下，**快速、频繁查询某个区间的累加和**。
 
 核心思路：开辟新数组 `preSum[i]` 来存储原数组 `nums[0..i-1] `的累加和，`preSum[0] = 0`。这样，当求原数组区间和就比较容易了，区间 `[i..j]` 的和等于 `preSum[j+1] - preSum[i]` 的结果值。
 
 ```JavaScript
 const preSum = [0]
-
 preSum[i] = preSum[i - 1] + nums[i - 1]
+// 查询
+preSum[r + 1] - preSum[l]
+
+// 或者
+const preSum = [nums[0]]
+preSum[i] = preSum[i-1] + nums[i]
+// 查询
+preSum[r] - preSum[l-1] // 需要判断 l 为 0 的情况，直接返回 preSum[r]，这种的劣势是需要判断边界条件
 ```
 
 ### 构造：[303.区域和检索-数组不可变](https://leetcode.cn/problems/range-sum-query-immutable/)
@@ -25,8 +32,8 @@ preSum[i] = preSum[i - 1] + nums[i - 1]
  * @param {number[]} nums
  */
 var NumArray = function(nums) {
-    this.preSum = [0] // presum 首位为0 便于计算
-    for(let i = 1; i <= nums.length; ++i) { // 遍历数组的每一个元素, 因为使用的是nums[i-1]
+    this.preSum = [0] // preSum 首位为0 便于计算
+    for(let i = 1; i <= nums.length; ++i) {
         this.preSum[i] = this.preSum[i - 1] + nums[i -1]
     }
 };
@@ -47,40 +54,78 @@ NumArray.prototype.sumRange = function(left, right) {
  */
 ```
 
+---
+
+上方那种事官解给出的逻辑，下面这种可能更容易理解一些。
+
+```js
+/**
+ * @param {number[]} nums
+ */
+var NumArray = function (nums) {
+  this.preSums = [nums[0]]
+  for (let i = 1; i < nums.length; ++i) {
+    this.preSums[i] = this.preSums[i - 1] + nums[i]
+  }
+}
+
+/**
+ * @param {number} left
+ * @param {number} right
+ * @return {number}
+ */
+NumArray.prototype.sumRange = function (left, right) {
+  if (left === 0) return this.preSums[right]
+  return this.preSums[right] - this.preSums[left - 1]
+}
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * var obj = new NumArray(nums)
+ * var param_1 = obj.sumRange(left,right)
+ */
+```
+
 ### 构造：[304. 二维区域和检索-矩阵不可变](https://leetcode.cn/problems/range-sum-query-2d-immutable/)
 
 ```js
 /**
  * @param {number[][]} matrix
  */
 var NumMatrix = function (matrix) {
-  const m = matrix.length,
-    n = matrix[0].length
-  if (m == 0 || n == 0) return
-  // 第一行,第一列填充为0方便计算
-  this.presum = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0))
-  // 遍历presum,因为要填充到最后一个,所以是闭区间,使用小于等于，所以前缀和的索引超前matrix 1位
-  for (let i = 1; i <= m; ++i) {
-    for (let j = 1; j <= n; ++j) {
-      // 计算矩阵[]0,0,i,j]，减去部分为重复部分
-      this.presum[i][j] =
-        this.presum[i - 1][j] +
-        this.presum[i][j - 1] -
-        this.presum[i - 1][j - 1] +
-        matrix[i - 1][j - 1]
-      // 当前块的区域和等于以matrix[0,0]为定点四个区域块的加减得来,其中左上角有一块会被加两次所以要减掉一次
+  const row = matrix.length
+  const col = matrix[0].length
+  this.sums = Array.from(Array(row + 1), () => Array(col + 1).fill(0))
+  for (let i = 0; i < row; ++i) {
+    for (let j = 0; j < col; ++j) {
+      this.sums[i + 1][j + 1] =
+        this.sums[i + 1][j] + this.sums[i][j + 1] - this.sums[i][j] + matrix[i][j]
     }
   }
+  console.log(this.sums)
 }
 
+/**
+ * @param {number} row1
+ * @param {number} col1
+ * @param {number} row2
+ * @param {number} col2
+ * @return {number}
+ */
 NumMatrix.prototype.sumRegion = function (row1, col1, row2, col2) {
   return (
-    this.presum[row2 + 1][col2 + 1] +
-    this.presum[row1][col1] -
-    this.presum[row1][col2 + 1] -
-    this.presum[row2 + 1][col1]
+    this.sums[row2 + 1][col2 + 1] -
+    this.sums[row1][col2 + 1] -
+    this.sums[row2 + 1][col1] +
+    this.sums[row1][col1]
   )
 }
+
+/**
+ * Your NumMatrix object will be instantiated and called as such:
+ * var obj = new NumMatrix(matrix)
+ * var param_1 = obj.sumRegion(row1,col1,row2,col2)
+ */
 ```
 
 ### 应用：[560.和为 k 的子数组](https://leetcode.cn/problems/subarray-sum-equals-k/)