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0303.SumRange.go
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0303.SumRange.go
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package _0303
/*
303. 区域和检索 - 数组不可变
给定一个整数数组 nums,求出数组从索引 i 到 j(i ≤ j)范围内元素的总和,包含 i、j 两点。
实现 NumArray 类:
NumArray(int[] nums) 使用数组 nums 初始化对象
int sumRange(int i, int j) 返回数组 nums 从索引 i 到 j(i ≤ j)范围内元素的总和,包含 i、j 两点(也就是 sum(nums[i], nums[i + 1], ... , nums[j]))
示例:
输入:
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出:
[null, 1, -1, -3]
解释:
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
提示:
0 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
最多调用 104 次 sumRange 方法
*/
type NumArray struct {
sum []int
}
func Constructor(nums []int) NumArray {
s := make([]int, len(nums)+1)
for i := 0; i < len(nums); i++ {
s[i+1] = s[i] + nums[i]
}
return NumArray{s}
}
func (n *NumArray) SumRange(i int, j int) int {
return n.sum[j+1] - n.sum[i]
}
/**
* Your NumArray object will be instantiated and called as such:
* obj := Constructor(nums);
* param_1 := obj.SumRange(i,j);
*/