leetcode-day4:Minimum Size Subarray Sum(长度最小的子数组)

209. Minimum Size Subarray Sum
数组：滑动窗口拯救了你

Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, …, numsr-1, numsr] of which the sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 105

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)).

JAVA实现代码：

/**
暴力解法：
1.双层循环，外层控制子数组起始值
2.内层循环从起始位置向后累加，直到和大于或者等于target，记录长度，跳出内层循环
3.外层数组起始值增加，不断测试更小长度
时间复杂度：O(n^2)

滑动窗口法：
1.双指针从起点开始，慢指针不动，快指针后移到两指针和大于等于target时，记录此时子数组长度
2.并从和中减去慢指针的值，慢指针向后移动，不断尝试更小的长度，一直到子数组和不再满足条件
3.慢指针继续保持不动，快指针后移，重复步骤2
时间复杂度：O(n)
*/
class Solution {
    public int minSubArrayLen(int target, int[] nums) {
		int sum=0;
		int slowIndex=0;
        int length=Integer.MAX_VALUE;
	
		for(int fastIndex=0;fastIndex<nums.length;fastIndex++){
            sum+=nums[fastIndex];
            while(sum>=target){
                int sumLength=fastIndex-slowIndex+1;
                length=sumLength<length?sumLength:length;
                sum-=nums[slowIndex++];
            }
		}
	
		return length==Integer.MAX_VALUE?0:length;
    }
}