[LeetCode] 209. Minimum Size Subarray Sum

轉自LeetCode

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). 

<Solution>

想法如下

• 用一個變數計算目前的 sum，當 sum >= s 的時候，開始找答案

• 當 sum >= s 這個條件成立時，計算出目前的 subarray 長度，然後縮起始位置並減掉該位置的值，看看 sum >= s 這個條件是不是還是成立，是的話，再重複做之前的動作

code 如下

Java(參考解法)

	class Solution {
	public int minSubArrayLen(int s, int[] nums) {
	int sum = 0, left = 0, ans = Integer.MAX_VALUE;
	for(int i = 0; i < nums.length; i++) {
	sum += nums[i];
	while(sum >= s) {
	ans = Math.min(ans, i - left + 1);
	sum -= nums[left++];
	}
	}
	return ans == Integer.MAX_VALUE ? 0 : ans;
	}
	}

view raw minSubArrayLen.java hosted with ❤ by GitHub

kotlin

	class Solution {
	fun minSubArrayLen(target: Int, nums: IntArray): Int {
	var ans = Int.MAX_VALUE
	var left = 0
	var sum = 0
	for(right in nums.indices) {
	sum += nums[right]
	while(sum >= target) {
	ans = Math.min(ans, right-left+1)
	sum -= nums[left++]
	}
	}
	return if(ans == Int.MAX_VALUE) 0 else ans
	}
	}

view raw minimum_size_subarray_sum.kt hosted with ❤ by GitHub