LeetCode 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.

分析

步骤

• 我们设置bottom和top控制子数组的头部和尾部。
• 初始化bottom=0，top为-1，表示一个空的子数组
• 子数组和sum=0，最小长度len=0。
• 当sum < s时，在当前子数组的尾部增加一个元素bottom[++top]。
• 当sum ≥ s时，去掉当前子数组的头部元素，并++bottom。
• 退出循环的条件：top == nums.size() 或 bottom>top（此时已经存在最小len为1，不可能更小，可以退出）。

AC代码

O(n)及O(nlogn)算法

``````//O(n)
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
if (!nums.size()) return 0;
int bottom = 0, top = -1;
int sum = 0, len = 0;
while (true) {
if (sum < s) {
++top;
if (top != nums.size())
sum += nums[top];
else
break;
} else {
sum -= nums[bottom]; ++bottom;
if (bottom > top)
break;
}
if (sum >= s) {
int new_len = top - bottom + 1;
if (!len || len && new_len < len)
len = new_len;
}
}
return len;
}
};

//O(nlogn)
class Solution {
public:
int MAX_INT = 999999999;
int minSubArrayLen(int s, vector<int>& nums) {
if (!nums.size()) return 0;
return findMinLen(s, nums, 0, nums.size() - 1);
}
int findMinLen(int s, vector<int>& nums, int bottom, int top) {
if (top == bottom) return nums[top] >= s ? 1 : 0;
int mid = (bottom + top) / 2;
int left = mid, right = mid + 1, sum = 0, len;
while (sum < s && (right <= top || left >= bottom)) {
if  (right > top) {
sum += nums[left]; --left;
}
else if (left < bottom) {
sum += nums[right]; ++right;
}
else if (nums[left] > nums[right]) {
sum += nums[left]; --left;
}
else {
sum += nums[right]; ++right;
}
}
if (sum >= s) {
len = right - left - 1;
int leftLen = findMinLen(s, nums, bottom, mid);
int rightLen = findMinLen(s, nums, mid + 1, top);
return minValue(leftLen, rightLen, len);
}
else {
return 0;
}
}
int minValue(int x, int y, int z) {
if (!x) x = MAX_INT;
if (!y) y = MAX_INT;
if (x <= y && x <= z) return x;
if (y <= x && y <= z) return y;
return z;
}
};
``````