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 1:
Input:nums = [1,1,1], k = 2
Output: 2
Note:
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].
题意:找连续子序列的和等于k的 子串个数
暴力解法:双重循环,以每个点为起点双重遍历一遍, O(n^2)
public static int subarraySum(int[] nums, int k) {
if (nums == null)
return 0;
int count = 0,left = 0,right = 0 ,cur = 0;
for (int i = 0 ; i < nums.length ;i++)
{
for (int j = i ; j <nums.length ; j++)
{
cur += nums[j];
if (cur == k)
count++;
}
cur = 0;
}
return count;
}
优化:要想计算 s[i,j] (i到j的sum)时,可以通过记忆化缓存之前的s[0,i]和s[0,j] 然后通过s[0,j]-s[0,i] = s[i,j]获得。那实际上当我们遍历的时候所有s[0,cur] (cur表示当前下标)其实都已经计算过了,只需要存下来就好了。
那么通过一次遍历到j的时候 我们就可以用hashmap 记录前面所有s[0,j]的值(value为这样的前缀的个数)。然后只要map中存在当前sum-target就说明 存在s[0,i]==sum - target(这里的sum当然也就是s[0,j]),也就是存在s[i,j]=target
public static int bettersubarraySum(int[] nums, int k) {
if (nums == null)
return 0;
int count = 0,sum = 0;
Map<Integer,Integer> map = new HashMap<>();
map.put(0,1); //初始化,值为0的前缀 个数为1
for (int i = 0 ; i < nums.length ;i++)
{
sum += nums[i];
if (map.containsKey(sum-k))
count += map.get(sum-k);
map.put(sum,map.getOrDefault(sum,0)+1);
}
return count;
}