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].
題意:找連續子序列的和等於k的 子串個數
暴力解法:雙重循環,以每個點爲起點雙重遍歷一遍, 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;
}
優化:要想計算 s[i,j] (i到j的sum)時,可以通過記憶化緩存之前的s[0,i]和s[0,j] 然後通過s[0,j]-s[0,i] = s[i,j]獲得。那實際上當我們遍歷的時候所有s[0,cur] (cur表示當前下標)其實都已經計算過了,只需要存下來就好了。
那麼通過一次遍歷到j的時候 我們就可以用hashmap 記錄前面所有s[0,j]的值(value爲這樣的前綴的個數)。然後只要map中存在當前sum-target就說明 存在s[0,i]==sum - target(這裏的sum當然也就是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,值爲0的前綴 個數爲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;
}