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560-SubarraySumEqualsK.go
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74 lines (64 loc) · 2.03 KB
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package main
// 560. Subarray Sum Equals K
// Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k.
// A subarray is a contiguous non-empty sequence of elements within an array.
// Example 1:
// Input: nums = [1,1,1], k = 2
// Output: 2
// Example 2:
// Input: nums = [1,2,3], k = 3
// Output: 2
// Constraints:
// 1 <= nums.length <= 2 * 10^4
// -1000 <= nums[i] <= 1000
// -10^7 <= k <= 10^7
import "fmt"
func subarraySum(nums []int, k int) int {
res, l := 0, len(nums)
for i := 0; i < l; i++ {
sum := 0
for j:= i; j < l; j++ {
sum += nums[j]
if sum == k {
res++
}
}
}
return res
}
// best solution
// 要求找到连续区间和为 k 的子区间总数,即区间 [i,j] 内的和为 K ⇒ prefixSum[j] - prefixSum[i-1] == k。
// 所以 prefixSum[j] == k - prefixSum[i-1]
func subarraySum1(nums []int, k int) int {
res, pre := 0,0
m := map[int]int{}
m[0] = 1
for i:= 0; i <len(nums); i++ {
pre += nums[i]
if _,ok := m[pre - k]; ok {
res += m[pre - k]
}
// 用 map 存储累加过的结果
m[pre] += 1
}
return res
}
func main() {
// Example 1:
// Input: nums = [1,1,1], k = 2
// Output: 2
fmt.Println(subarraySum([]int{1,1,1}, 2)) // 2
// Example 2:
// Input: nums = [1,2,3], k = 3
// Output: 2
fmt.Println(subarraySum([]int{1,2,3}, 3)) // 2
fmt.Println(subarraySum([]int{1,2,3,4,5,6,7,8,0}, 22)) // 1
fmt.Println(subarraySum([]int{1,2,3,4,5,6,7,8,0}, 22)) // 1
fmt.Println(subarraySum([]int{1,2,3,4,5,6,7,8,9}, 2)) // 1
fmt.Println(subarraySum([]int{9,8,7,6,5,4,3,2,1}, 2)) // 1
fmt.Println(subarraySum1([]int{1,1,1}, 2)) // 2
fmt.Println(subarraySum1([]int{1,2,3}, 3)) // 2
fmt.Println(subarraySum1([]int{1,2,3,4,5,6,7,8,0}, 22)) // 1
fmt.Println(subarraySum1([]int{1,2,3,4,5,6,7,8,9}, 2)) // 1
fmt.Println(subarraySum1([]int{9,8,7,6,5,4,3,2,1}, 2)) // 1
}