Subarray Sum Equals K
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Example
Input: nums = [1,1,1], k = 2
Output: 2
Input: nums = [1,2,-1,-1,2], k = 2
Output: 4
The length of the array is in range [1, 20,000].
The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
Note
Using HashMap to store ( key : the PrefixSum, value : how many ways get to this prefix sum).
Check if PrefixSum - target exists in hashmap or not, if it does, we added up the ways of this key
For instance : in one path we have 1,2,-1,-1,2, then the prefix sum will be: 0, 1, 3, 2, 1, 3, let's say we want to find target sum is 2, then we will have{2}, {1,2,-1}, {2,-1,-1,2} and {2}ways.
Meaning:
PrefixSum2 - PrefixSum1 = Target
注意One Pass写法时候,其实没有把起始0加入循环,需要手动设置成map.put(0,1). PreSum数组则不用,因为它从index 0遍历到index len + 1。
Code
class Solution {
public int subarraySum(int[] nums, int k) {
int len = nums.length;
int[] presum = new int[len + 1];
for (int i = 0; i < len; i++) {
presum[i + 1] = presum[i] + nums[i];
}
int res = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i <= len; i++) {
int key = presum[i] - k;
if (map.containsKey(key)) {
res += map.get(key);
}
map.put(presum[i], map.getOrDefault(presum[i], 0) + 1);
}
return res;
}
}
class Solution {
public int subarraySum(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
map.put(0, 1);
int res = 0, sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (map.containsKey(sum - k)) {
res += map.get(sum - k);
}
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
return res;
}
}
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