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.
思路：用start, end两个游标来记录范围，sum < s end就向后走， s >= sum start就向后走。
我写的代码没有大神的逻辑清晰，先上大神的。
int minSubArrayLen(int s, vector<int>& nums) {
int firstPos = , sum = , minLength = INT_MAX;
for(int i = ; i<nums.size(); i++) { //i即end游标 对所有end游标循环
sum += nums[i];
while(sum >= s) { //对每个end游标的start游标循环 firstPos即为start游标 只有s >= sum 时才把start向后移
minLength = min(minLength, i - firstPos + );
sum -= nums[firstPos++];
}
}
return minLength == INT_MAX? : minLength; //没找到s >= sum 时返回0
}
我的代码乱一点，但是也AC了。
int minSubArrayLen(int s, vector<int>& nums) {
int start = , end = ;
int sum = ;
int minLength = nums.size() + ;
while(end <= nums.size()) //有等于是因为结尾到最后面时 起始点还可能移动
{
if(sum < s)
{
if(end == nums.size()) break;
sum += nums[end++];
}
else
{
minLength = (minLength < (end - start)) ? minLength : (end - start);
sum -= nums[start++];
}
}
minLength = (minLength == nums.size() + ) ? : minLength; //没找到符合条件的子序列 返回0
return minLength;
}