[LeetCode] 209. Minimum Size Subarray Sum 最短子数组之和

Posted 2021-01-07 轻风舞动

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). 

这道题的关键是由正整数组成的数组，这样才能保证累加的数组是递增的，才能使用双指针或者二分法。

解法1: 双指针，滑动窗口。

解法2: 二分法

Java: moving window

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

Java: BS

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int i = 1, j = nums.length, min = 0;
        while (i <= j) {
            int mid = (i + j) / 2;
            if (windowExist(mid, nums, s)) {
                j = mid - 1;
                min = mid;
            } else i = mid + 1;
        }
        return min;
    }


    private boolean windowExist(int size, int[] nums, int s) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i >= size) sum -= nums[i - size];
            sum += nums[i];
            if (sum >= s) return true;
        }
        return false;
    }
}

Java: BS　　

public class Solution {
 public int minSubArrayLen(int s, int[] nums) {
        int sum = 0, min = Integer.MAX_VALUE;

        int[] sums = new int[nums.length];
        for (int i = 0; i < nums.length; i++)
            sums[i] = nums[i] + (i == 0 ? 0 : sums[i - 1]);

        for (int i = 0; i < nums.length; i++) {
            int j = findWindowEnd(i, sums, s);
            if (j == nums.length) break;
            min = Math.min(j - i + 1, min);
        }
        
        return min == Integer.MAX_VALUE ? 0 : min;
    }

    private int findWindowEnd(int start, int[] sums, int s) {
        int i = start, j = sums.length - 1, offset = start == 0 ? 0 : sums[start - 1];
        while (i <= j) {
            int m = (i + j) / 2;
            int sum = sums[m] - offset;
        if (sum >= s) j = m - 1;
        else i = m + 1;
    }
    return i;
}　　　

Java:

public int minSubArrayLen(int s, int[] a) {
  if (a == null || a.length == 0)
    return 0;
  
  int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
  
  while (j < a.length) {
    sum += a[j++];
    
    while (sum >= s) {
      min = Math.min(min, j - i);
      sum -= a[i++];
    }
  }
  
  return min == Integer.MAX_VALUE ? 0 : min;
}

Java:

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        return solveNLogN(s, nums);
    }
    
    private int solveN(int s, int[] nums) {
        int start = 0, end = 0, sum = 0, minLen = Integer.MAX_VALUE;
        while (end < nums.length) {
            while (end < nums.length && sum < s) sum += nums[end++];
            if (sum < s) break;
            while (start < end && sum >= s) sum -= nums[start++];
            if (end - start + 1 < minLen) minLen = end - start + 1;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }

    private int solveNLogN(int s, int[] nums) {
        int[] sums = new int[nums.length + 1];
        for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
        int minLen = Integer.MAX_VALUE;
        for (int i = 0; i < sums.length; i++) {
            int end = binarySearch(i + 1, sums.length - 1, sums[i] + s, sums);
            if (end == sums.length) break;
            if (end - i < minLen) minLen = end - i;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }
    
    private int binarySearch(int lo, int hi, int key, int[] sums) {
        while (lo <= hi) {
           int mid = (lo + hi) / 2;
           if (sums[mid] >= key){
               hi = mid - 1;
           } else {
               lo = mid + 1;
           }
        }
        return lo;
    }
}　

Python:

# Sliding window solution.
class Solution:
    # @param {integer} s
    # @param {integer[]} nums
    # @return {integer}
    def minSubArrayLen(self, s, nums):
        start = 0
        sum = 0
        min_size = float("inf")
        for i in xrange(len(nums)):
            sum += nums[i]
            while sum >= s:
                min_size = min(min_size, i - start + 1)
                sum -= nums[start]
                start += 1

        return min_size if min_size != float("inf") else 0

Python:　　

# Time:  O(nlogn)
# Space: O(n)
# Binary search solution.
class Solution2:
    # @param {integer} s
    # @param {integer[]} nums
    # @return {integer}
    def minSubArrayLen(self, s, nums):
        min_size = float("inf")
        sum_from_start = [n for n in nums]
        for i in xrange(len(sum_from_start) - 1):
            sum_from_start[i + 1] += sum_from_start[i]
        for i in xrange(len(sum_from_start)):
            end = self.binarySearch(lambda x, y: x <= y, sum_from_start, \\
                                    i, len(sum_from_start), \\
                                    sum_from_start[i] - nums[i] + s)
            if end < len(sum_from_start):
                min_size = min(min_size, end - i + 1)

        return min_size if min_size != float("inf") else 0

    def binarySearch(self, compare, A, start, end, target):
        while start < end:
            mid = start + (end - start) / 2
            if compare(target, A[mid]):
                end = mid
            else:
                start = mid + 1
        return start　　

C++:

/ O(n)
class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if (nums.empty()) return 0;
        int left = 0, right = 0, sum = 0, len = nums.size(), res = len + 1;
        while (right < len) {
            while (sum < s && right < len) {
                sum += nums[right++];
            }
            while (sum >= s) {
                res = min(res, right - left);
                sum -= nums[left++];
            }
        }
        return res == len + 1 ? 0 : res;
    }
};　　

C++:

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int res = INT_MAX, left = 0, sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            while (left <= i && sum >= s) {
                res = min(res, i - left + 1);
                sum -= nums[left++];
            }
        }
        return res == INT_MAX ? 0 : res;
    }
};

C++: 二分法

// O(nlgn)
class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int len = nums.size(), sums[len + 1] = {0}, res = len + 1;
        for (int i = 1; i < len + 1; ++i) sums[i] = sums[i - 1] + nums[i - 1];
        for (int i = 0; i < len + 1; ++i) {
            int right = searchRight(i + 1, len, sums[i] + s, sums);
            if (right == len + 1) break;
            if (res > right - i) res = right - i;
        }
        return res == len + 1 ? 0 : res;
    }
    int searchRight(int left, int right, int key, int sums[]) {
        while (left <= right) {
            int mid = (left + right) / 2;
            if (sums[mid] >= key) right = mid - 1;
            else left = mid + 1;
        }
        return left;
    }
};

C++:

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int res = INT_MAX, n = nums.size();
        vector<int> sums(n + 1, 0);
        for (int i = 1; i < n + 1; ++i) sums[i] = sums[i - 1] + nums[i - 1];
        for (int i = 0; i < n; ++i) {
            int left = i + 1, right = n, t = sums[i] + s;
            while (left <= right) {
                int mid = left + (right - left) / 2;
                if (sums[mid] < t) left = mid + 1;
                else right = mid - 1;
            }
            if (left == n + 1) break;
            res = min(res, left - i);
        }
        return res == INT_MAX ? 0 : res;
    }
};

 

 

类似题目：

[LeetCode] 53. Maximum Subarray 最大子数组

[LeetCode] 325. Maximum Size Subarray Sum Equals k 和等于k的最长子数组

[LeetCode] 560. Subarray Sum Equals K 子数组和为K

 

All LeetCode Questions List 题目汇总