303

Author: wind-liang

SUMMARY.md

@@ -247,4 +247,5 @@
     * [300. Longest Increasing Subsequence](leetcode-300-Longest-Increasing-Subsequence.md)
 * [301 题到 400 题](leetcode-301-400.md)    
     * [301. Remove Invalid Parentheses](leetcode-301-Remove-Invalid-Parentheses.md)
+    * [303. Range Sum Query - Immutable](leetcode-303-Range-Sum-Query-Immutable.md)
 * [更多](more.md)

leetcode-303-Range-Sum-Query-Immutable.md

@@ -0,0 +1,141 @@
+# 题目描述（简单难度）
+
+303、Range Sum Query - Immutable
+
+Given an integer array *nums*, find the sum of the elements between indices *i* and *j* (*i* ≤ *j*), inclusive.
+
+**Example:**
+
+```
+Given nums = [-2, 0, 3, -5, 2, -1]
+
+sumRange(0, 2) -> 1
+sumRange(2, 5) -> -1
+sumRange(0, 5) -> -3
+```
+
+
+
+**Note:**
+
+1. You may assume that the array does not change.
+2. There are many calls to *sumRange* function.
+
+设计一个类，返回数组的第 `i` 到 `j` 个元素的和。
+
+# 解法一 暴力
+
+题目是想让我们设计一种数据结构，我们先暴力写一下。
+
+```javascript
+/**
+ * @param {number[]} nums
+ */
+var NumArray = function(nums) {
+    this.nums = nums;
+};
+
+/** 
+ * @param {number} i 
+ * @param {number} j
+ * @return {number}
+ */
+NumArray.prototype.sumRange = function(i, j) {
+    let sum = 0;
+    for(; i <= j; i++){
+        sum += this.nums[i];
+    }
+    return sum;
+};
+
+/** 
+ * Your NumArray object will be instantiated and called as such:
+ * var obj = new NumArray(nums)
+ * var param_1 = obj.sumRange(i,j)
+ */
+```
+
+分享 [官方](https://leetcode.com/problems/range-sum-query-immutable/solution/) 提供的一个优化，因为是多次调用 `sumRange`，我们可以在每次调用后将结果保存起来，这样的话第二次调用就可以直接返回了。
+
+```java
+/**
+ * @param {number[]} nums
+ */
+var NumArray = function(nums) {
+    this.nums = nums;
+    this.map = {};
+};
+
+/** 
+ * @param {number} i 
+ * @param {number} j
+ * @return {number}
+ */
+NumArray.prototype.sumRange = function(i, j) {
+    let key = i + '@' + j;
+    if(this.map.hasOwnProperty(key)){
+        return this.map[key];
+    }
+    
+    let sum = 0;
+    for(; i <= j; i++){
+        sum += this.nums[i];
+    }
+    
+    this.map[key] = sum;
+    return sum;
+};
+
+/** 
+ * Your NumArray object will be instantiated and called as such:
+ * var obj = new NumArray(nums)
+ * var param_1 = obj.sumRange(i,j)
+ */
+```
+
+# 解法二
+
+和 [238 题](https://leetcode.wang/leetcode-238-Product-of-Array-Except-Self.html) 的解法二有一些像。
+
+我们用一个数组保存累计的和，`numsAccumulate[i]` 存储 `0` 到 `i - 1` 累计的和。
+
+如果我们想求 `i` 累积到 `j` 的和，只需要用 `numsAccumulate[j + 1]` 减去 `numsAccumulate[i]`。
+
+结合下边的图应该很好理解，我们要求的是橙色部分，相当于 `B` 的部分减去 `A` 的部分。
+
+![](https://windliang.oss-cn-beijing.aliyuncs.com/303_2.jpg)
+
+代码也就水到渠成了。
+
+```javascript
+/**
+ * @param {number[]} nums
+ */
+var NumArray = function (nums) {
+    this.numsAccumulate = [0];
+    let sum = 0;
+    for (let i = 0; i < nums.length; i++) {
+        sum += nums[i];
+        this.numsAccumulate.push(sum);
+    }
+};
+
+/**
+ * @param {number} i
+ * @param {number} j
+ * @return {number}
+ */
+NumArray.prototype.sumRange = function (i, j) {
+    return this.numsAccumulate[j + 1] - this.numsAccumulate[i];
+};
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * var obj = new NumArray(nums)
+ * var param_1 = obj.sumRange(i,j)
+ */
+```
+
+# 总
+
+比较简单的一道题，解法一做缓存的思路比较常见。解法二的思路印象中也遇到过几次，看到题目我的第一反应想到的就是解法二。