209. Minimum Size Subarray Sum
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.
这个题目一来就应该想到应该一点一点累加恰好找到这个s的累加和,统计出长度
所以接着缩小数据,再次找到恰好大于s的位置,更新最小长度,恰好是为了尽可能小
具体:用一个迭代器慢ite1,一个迭代器快ite2
如果当前序列的和小于s则ite2继续右移动(增加和),否则ite1右移动(相当于减小和)
class Solution {
public:
int minSubArrayLen(int s, vector& nums) {
if(nums.empty())
return 0;
vector::iterator ite1=nums.begin();
vector::iterator ite2=ite1;
int minLen=INT_MAX;
int curSum=*ite1,curLen=1;
while(true)
{
if(curSum >= s )
{
if(curLen
别人的算法
(本质一样),O(N)的速度:
class Solution {
public:
int minSubArrayLen(int s, vector& nums) {
int n = nums.size(), start = 0, sum = 0, minlen = INT_MAX;
for (int i = 0; i < n; i++) {
sum += nums[i];
while (sum >= s) {
minlen = min(minlen, i - start + 1);
sum -= nums[start++];
}
}
return minlen == INT_MAX ? 0 : minlen;
}
};
注:本博文为EbowTang原创,后续可能继续更新本文。如果转载,请务必复制本条信息!
原文地址:http://blog.csdn.net/ebowtang/article/details/50453943
原作者博客:http://blog.csdn.net/ebowtang