|
| 1 | +#题目描述（简单难度） |
| 2 | + |
| 3 | +303、Range Sum Query - Immutable |
| 4 | + |
| 5 | +Given an integer array*nums*, find the sum of the elements between indices*i* and*j* (*i* ≤*j*), inclusive. |
| 6 | + |
| 7 | +**Example:** |
| 8 | + |
| 9 | +``` |
| 10 | +Given nums = [-2, 0, 3, -5, 2, -1] |
| 11 | +
|
| 12 | +sumRange(0, 2) -> 1 |
| 13 | +sumRange(2, 5) -> -1 |
| 14 | +sumRange(0, 5) -> -3 |
| 15 | +``` |
| 16 | + |
| 17 | + |
| 18 | + |
| 19 | +**Note:** |
| 20 | + |
| 21 | +1. You may assume that the array does not change. |
| 22 | +2. There are many calls to*sumRange* function. |
| 23 | + |
| 24 | +设计一个类，返回数组的第`i` 到`j` 个元素的和。 |
| 25 | + |
| 26 | +#解法一 暴力 |
| 27 | + |
| 28 | +题目是想让我们设计一种数据结构，我们先暴力写一下。 |
| 29 | + |
| 30 | +```javascript |
| 31 | +/** |
| 32 | + *@param{number[]}nums |
| 33 | +*/ |
| 34 | +varNumArray=function(nums) { |
| 35 | +this.nums= nums; |
| 36 | +}; |
| 37 | + |
| 38 | +/** |
| 39 | + *@param{number}i |
| 40 | + *@param{number}j |
| 41 | + *@return{number} |
| 42 | +*/ |
| 43 | +NumArray.prototype.sumRange=function(i,j) { |
| 44 | +let sum=0; |
| 45 | +for(; i<= j; i++){ |
| 46 | + sum+=this.nums[i]; |
| 47 | + } |
| 48 | +return sum; |
| 49 | +}; |
| 50 | + |
| 51 | +/** |
| 52 | + * Your NumArray object will be instantiated and called as such: |
| 53 | + * var obj = new NumArray(nums) |
| 54 | + * var param_1 = obj.sumRange(i,j) |
| 55 | +*/ |
| 56 | +``` |
| 57 | + |
| 58 | +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。 |
| 59 | + |
| 60 | +```java |
| 61 | +/** |
| 62 | + *@param {number[]} nums |
| 63 | +*/ |
| 64 | +varNumArray= function(nums) { |
| 65 | +this.nums= nums; |
| 66 | +this.map= {}; |
| 67 | +}; |
| 68 | + |
| 69 | +/** |
| 70 | + *@param {number} i |
| 71 | + *@param {number} j |
| 72 | + *@return {number} |
| 73 | +*/ |
| 74 | +NumArray.prototype.sumRange= function(i, j) { |
| 75 | + let key= i+'@'+ j; |
| 76 | +if(this.map.hasOwnProperty(key)){ |
| 77 | +returnthis.map[key]; |
| 78 | + } |
| 79 | + |
| 80 | + let sum=0; |
| 81 | +for(; i<= j; i++){ |
| 82 | + sum+=this.nums[i]; |
| 83 | + } |
| 84 | + |
| 85 | +this.map[key]= sum; |
| 86 | +return sum; |
| 87 | +}; |
| 88 | + |
| 89 | +/** |
| 90 | + * Your NumArray object will be instantiated and called as such: |
| 91 | + * var obj = new NumArray(nums) |
| 92 | + * var param_1 = obj.sumRange(i,j) |
| 93 | +*/ |
| 94 | +``` |
| 95 | + |
| 96 | +#解法二 |
| 97 | + |
| 98 | +和[238 题](https://leetcode.wang/leetcode-238-Product-of-Array-Except-Self.html) 的解法二有一些像。 |
| 99 | + |
| 100 | +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。 |
| 101 | + |
| 102 | +如果我们想求`i` 累积到`j` 的和，只需要用`numsAccumulate[j + 1]` 减去`numsAccumulate[i]`。 |
| 103 | + |
| 104 | +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。 |
| 105 | + |
| 106 | + |
| 107 | + |
| 108 | +代码也就水到渠成了。 |
| 109 | + |
| 110 | +```javascript |
| 111 | +/** |
| 112 | + *@param{number[]}nums |
| 113 | +*/ |
| 114 | +varNumArray=function (nums) { |
| 115 | +this.numsAccumulate= [0]; |
| 116 | +let sum=0; |
| 117 | +for (let i=0; i<nums.length; i++) { |
| 118 | + sum+= nums[i]; |
| 119 | +this.numsAccumulate.push(sum); |
| 120 | + } |
| 121 | +}; |
| 122 | + |
| 123 | +/** |
| 124 | + *@param{number}i |
| 125 | + *@param{number}j |
| 126 | + *@return{number} |
| 127 | +*/ |
| 128 | +NumArray.prototype.sumRange=function (i,j) { |
| 129 | +returnthis.numsAccumulate[j+1]-this.numsAccumulate[i]; |
| 130 | +}; |
| 131 | + |
| 132 | +/** |
| 133 | + * Your NumArray object will be instantiated and called as such: |
| 134 | + * var obj = new NumArray(nums) |
| 135 | + * var param_1 = obj.sumRange(i,j) |
| 136 | +*/ |
| 137 | +``` |
| 138 | + |
| 139 | +#总 |
| 140 | + |
| 141 | +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。 |