Description:
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.
Solution:
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
auto sz = (int)nums.size();
if (sz == 0) return 0;
int rc = 0;
int pre = 0;
int sum = 0;
int pos = 0;
while (pos < sz) {
sum += nums[pos];
if (sum < s) {
++pos;
} else if (sum == s) {
rc = (rc == 0) ? pos-pre+1 : min(rc, pos-pre+1);
sum -= nums[pre];
++pre;
++pos;
} else {
rc = (rc == 0) ? pos-pre+1 : min(rc, pos-pre+1);
sum = sum - nums[pre] - nums[pos];
++pre;
}
}
return rc;
}
};
转载于:https://www.cnblogs.com/deofly/p/minimum-size-subarray-sum.html
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