标签：compare vat poi spl subarray coding break als minimal
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.
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] 560. Subarray Sum Equals K 子数组和为K
[LeetCode] 209. Minimum Size Subarray Sum 最短子数组之和
标签：compare vat poi spl subarray coding break als minimal
原文地址：https://www.cnblogs.com/lightwindy/p/9678709.html
