LeetCode—Minimum Size Subarray Sum
2015-05-18 01:05
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题目:
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return
0 instead.
For example, given the array
the subarray
代码1:遍历,复杂度O(n^2)
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return
0 instead.
For example, given the array
[2,3,1,2,4,3]and
s = 7,
the subarray
[4,3]has the minimal length under the problem constraint.
代码1:遍历,复杂度O(n^2)
int min(int x,int y) { return x<y?x:y; } int minSubArrayLen(int s, int* nums, int numsSize) { int minlen = 100000; int count = 0; int i; int j; for(i = 0;i < numsSize;i ++) { count+=nums[i]; } if(count<s) return 0; count = 0; for(i = 0;i<numsSize;i++) { count = 0; for(j = i;j>=0;j--) { count += nums[j]; if(s <= count) { minlen = min(minlen,i-j+1); break; } } } return minlen; }C代码2:滑动窗口,复杂度O(n)
int minSubArrayLen(int s, int* nums, int numsSize) { int minlen = 100000; int left = 0; int right = 0; int sum = 0; while (right <= numsSize) { if (sum >= s) sum -= nums[left++]; else sum += nums[right++]; if (right - left < minlen && sum >= s) minlen = right - left; if(right == numsSize && sum < s) break; } return minlen == 100000 ? 0 : minlen; }复杂度为O(n*logn)的没想出来,thinking。。。
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