leetcode Minimum Size Subarray Sum java
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利用滑动窗口向右滑动并计算窗口的和 时间复杂度O(n) 空间复杂度(1)
public int minSubArrayLen(int s, int[] nums) {
if(nums==null||nums.length==0)return 0;
if(s==1)
return 1;
int start=0,end=0; //窗口起始点和结束点
int temp=0;
int cur=nums.length+1;
for(int i=0;i<nums.length;i++){
if(nums[i]>=s)
return 1;
temp+=nums[i];
if(temp>=s)
{
end=i;
if((end-start+1)<cur)
cur=end-start+1;
do{ //当前窗口和大于等于s,则起始点右移 再将新和与s比较
temp-=nums[start];
start++;
if((end-start+1)<cur&&temp>=s)
cur=end-start+1;
}while(temp>=s&&start<end);
}
}
return cur==nums.length+1?0:cur;
}
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