12
05/2015
[LeetCode] Minimum Size Subarray Sum
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.
解题思路：
注意这里的subarray是只顺序不可变，故不可以用先排序后选择的方法。
1、暴力滑动窗口法。依次验证窗口大小为1,2,3...的窗口，看看是否有和大于给定值的子数组。时间复杂度为O(n^2)，空间复杂度为1
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int size = nums.size();
int len = 1;
while(len<=size){
int sum = 0;
for(int j=0; j<len; j++){
sum += nums[j];
}
if(sum>=s){
return len;
}
int start = 0;
int end = start + len - 1;
while(end + 1<size){
sum -= nums[start];
sum += nums[end+1];
if(sum>=s){
return len;
}else{
start++;
end++;
}
}
len++;
}
return 0;
}
};2、伸缩窗口法。用两个指针start，end，若start和end-1之间的和小于s，end增加，若大于等于s，start增加，然后记录大于等于s时最小的长度。这种办法时间复杂度为O(n)，空间复杂度为1。
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int size = nums.size();
if(size==0){
return 0;
}
int minLen = size + 1;
int sum = 0;
int start=0, end=0;
while(end<size){
while(sum<s && end<size){
sum += nums[end];
end++;
}
while(sum>=s){
minLen = min(minLen, end - start);
if(minLen == 1){
return minLen;
}
sum -= nums[start];
start++;
}
}
return minLen==size + 1?0:minLen;
}
};
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