209. Minimum Size Subarray Sum

Medium

題目描述

Description

Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose 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] <= 104

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)).

解答

Solution

/**
 * @param {number} target
 * @param {number[]} nums
 * @return {number}
 */
var minSubArrayLen = function (target, nums) {
  let minLength = nums.length + 1;
  let left = 0;
  let right = 0;
  let currentSum = 0;

while (right < nums.length) {
currentSum += nums[right];

    while (currentSum >= target) {
      if (minLength > right - left + 1) {
        minLength = right - left + 1;
      }
      currentSum -= nums[left];
      left++;
    }

    right++;

}
if (minLength === nums.length + 1) {
return 0;
} else {
return minLength;
}
};

解題思路

練習 slide window，首先由 right 開始慢慢往右移動，並加總 left 和 right 範圍的這個 window 中的總和，直到大於 target。當大於 target 時候 left 就往右移動來減少 minLength 的可能值，重複移動 left 和 right 直到找出正確答案。

心得

也是看了 Wilson Ren 的 資料結構與演算法 (JavaScript) 介紹這題經典題，要把這些題型弄熟打好演算法基礎!