Minimum Size Subarray Sum (M)
Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.
Example:
Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
题意
给定一个正整数s和一个正整数数组nums,要求在nums中找到一个最短的连续子序列,使其和大于等于s。
思路
**滑动窗口法(Two Pointers):**连续子数组问题很容易想到使用滑动窗口来解题。用left和right分别指向窗口的左端点和右端点,如果当前窗口内的整数和小于s,则将右端点右移来拉长窗口;如果当前窗口内的整数和大于等于s,判断其长度是否比当前结果小,是则记录,并将左端点右移来缩短窗口。重复以上步骤就能得到一个长度最短且整数和大于等于s的窗口。最坏情况下左右端点都走了一遍数组,所以时间复杂度为O(2N)=O(N)O(2N)=O(N)O(2N)=O(N)。
**二分查找:**新建sum数组,sum[i]=nums[0]+...+nums[i]sum[i] = nums[0]+...+nums[i]sum[i]=nums[0]+...+nums[i],则问题转化为在数组sum中求一个连续子数组,使其末首之差大于等于s。从左向右遍历,每次固定左端点i,在右半边子数组中利用二分法找到值大于等于s+sum[i]s+sum[i]s+sum[i]的元素sum[j],判断长度并记录。时间复杂度为O(NlogN)O(NlogN)O(NlogN)。
代码实现 - 滑动窗口1
class Solution {
public int minSubArrayLen(int s, int[] nums) {
if (nums.length == 0) {
return 0;
}
int ans = 0;
int left = 0, right = 0;
int sum = nums[0];
while (right < nums.length) {
if (sum < s && right < nums.length - 1) {
sum += nums[++right];
} else if (sum >= s) {
ans = ans == 0 ? right - left + 1 : Math.min(ans, right - left + 1);
sum -= nums[left++];
} else {
break; // 当右端点已经在数组的末尾且sum仍比s小时,可以直接跳出
}
}
return ans;
}
}
代码实现 - 滑动窗口2
class Solution {
public int minSubArrayLen(int s, int[] nums) {
int ans = Integer.MAX_VALUE;
int sum = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
sum += nums[right];
while (sum >= s) {
ans = Math.min(ans, right - left + 1);
sum -= nums[left++];
}
}
return ans == Integer.MAX_VALUE ? 0 : ans;
}
}
代码实现 - 二分查找
class Solution {
public int minSubArrayLen(int s, int[] nums) {
if (nums.length == 0) {
return 0;
}
int ans = Integer.MAX_VALUE;
int[] sum = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
sum[i] = i == 0 ? nums[i] : sum[i - 1] + nums[i];
// 先处理掉特殊情况
if (sum[i] >= s) {
ans = Math.min(ans, i + 1);
}
}
// 二分查找符合条件的点
for (int i = 0; i < nums.length; i++) {
int target = s + sum[i];
int left = i + 1, right = nums.length - 1;
while (left < right) {
int mid = (right - left) / 2 + left;
if (sum[mid] < target) {
left = mid + 1;
} else {
ans = Math.min(ans, mid - i);
right = mid;
}
}
if (sum[right] >= target) {
ans = Math.min(ans, right - i);
}
}
return ans == Integer.MAX_VALUE ? 0 : ans;
}
}