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303.区域和检索-数组不可变.go

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+/*
+ * @lc app=leetcode.cn id=303 lang=golang
+ *
+ * [303] 区域和检索 - 数组不可变
+ *
+ * https://leetcode-cn.com/problems/range-sum-query-immutable/description/
+ *
+ * algorithms
+ * Easy (74.50%)
+ * Likes:    451
+ * Dislikes: 0
+ * Total Accepted:    162.3K
+ * Total Submissions: 217.4K
+ * Testcase Example:  '["NumArray","sumRange","sumRange","sumRange"]\n' +
+  '[[[-2,0,3,-5,2,-1]],[0,2],[2,5],[0,5]]'
+ *
+ * 给定一个整数数组  nums，处理以下类型的多个查询:
+ *
+ *
+ * 计算索引 left 和 right （包含 left 和 right）之间的 nums 元素的 和 ，其中 left <= right
+ *
+ *
+ * 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 中索引 left 和 right 之间的元素的 总和 ，包含 left 和
+ * right 两点（也就是 nums[left] + nums[left + 1] + ... + nums[right] )
+ *
+ *
+ *
+ *
+ * 示例 1：
+ *
+ *
+ * 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：
+ *
+ *
+ * 1 <= nums.length <= 10^4
+ * -10^5 <= nums[i] <= 10^5
+ * 0 <= i <= j < nums.length
+ * 最多调用 10^4 次 sumRange 方法
+ *
+ *
+*/
+
+// @lc code=start
+type NumArray struct {
+	PreSum []int
+}
+
+func Constructor(nums []int) NumArray {
+	preSum := make([]int, len(nums)+1)
+
+	// preSum[0] = 0
+	// preSum[1] = 0 + nums[0]
+	for i := 1; i <= len(nums); i++ {
+		preSum[i] = preSum[i-1] + nums[i-1]
+	}
+
+	return NumArray{
+		PreSum: preSum,
+	}
+}
+
+func (this *NumArray) SumRange(left int, right int) int {
+	return this.PreSum[right+1] - this.PreSum[left]
+
+}
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * obj := Constructor(nums);
+ * param_1 := obj.SumRange(left,right);
+ */
+// @lc code=end
+

560.和为-k-的子数组.go

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+/*
+ * @lc app=leetcode.cn id=560 lang=golang
+ *
+ * [560] 和为 K 的子数组
+ *
+ * https://leetcode-cn.com/problems/subarray-sum-equals-k/description/
+ *
+ * algorithms
+ * Medium (45.01%)
+ * Likes:    1491
+ * Dislikes: 0
+ * Total Accepted:    226.9K
+ * Total Submissions: 502.8K
+ * Testcase Example:  '[1,1,1]\n2'
+ *
+ * 给你一个整数数组 nums 和一个整数 k ，请你统计并返回 该数组中和为 k 的子数组的个数 。
+ * 
+ * 
+ * 
+ * 示例 1：
+ * 
+ * 
+ * 输入：nums = [1,1,1], k = 2
+ * 输出：2
+ * 
+ * 
+ * 示例 2：
+ * 
+ * 
+ * 输入：nums = [1,2,3], k = 3
+ * 输出：2
+ * 
+ * 
+ * 
+ * 
+ * 提示：
+ * 
+ * 
+ * 1 <= nums.length <= 2 * 10^4
+ * -1000 <= nums[i] <= 1000
+ * -10^7 <= k <= 10^7
+ * 
+ * 
+ */
+
+// @lc code=start
+func subarraySum(nums []int, k int) int {
+	pre, count := 0, 0
+	m := make(map[int]int, 0)
+	m[0] = 1
+	for i := range nums {
+		pre += nums[i]
+		if v, ok := m[pre-k];ok{
+			count += v
+		}
+		m[pre] += 1
+	}
+	return count
+}
+// @lc code=end
+