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].
刷出來之後居然還洋洋自得覺得自己真厲害
後來發現我寫了個o(n^2), 這和brute force有什麼區別!!空間複雜度分明可以O(1), 但是我居然用到了O(n), Shame on me!!
class Solution {
public int subarraySum(int[] nums, int k) {
if (nums == null || nums.length == 0) return k == 0 ? 1 : 0;
int[] res = new int[nums.length];
if (nums[0] == k) res[0] = 1;
else res[0] = 0;
for(int i = 1; i < nums.length; i++){
int count = 0;
int sum = 0;
for(int j = i; j >= 0; j--){
sum += nums[j];
if (sum == k) count++;
}
res[i] = res[i-1] + count;
}
return res[nums.length - 1];
}
}
如果要優化到比較好的O(n)的思路。
自然要從空間複雜度下手,那必須要存儲一些中間結果,那麼就是中間結果該怎麼保存的問題。
我是沒想明白,答案是這麼說的
Sum(i, j ) = Sum(0, j) - Sum(0, i-1)
哎,靠譜啊,再寫一版吧。
改良版本
class Solution {
public int subarraySum(int[] nums, int k) {
if (nums == null || nums.length == 0) return k == 0 ? 1 : 0;
Map<Integer, Integer> map = new HashMap<>();
int sum = 0;
int count = 0;
map.put(0, 1);
for(int i = 0; i < nums.length; i++){
sum += nums[i];
count += map.getOrDefault(sum - k, 0);
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
return count;
}
}
btw, 我煲的牛尾雜蔬湯可真好喝啊