description
Given an array containing n positive integers and a positive integer s, find the continuous sub-array with the smallest length satisfying its sum ≥ s and return its length. If there is no matching sub-array, 0 is returned.
Example:
Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: Sub-array [4,3] is the smallest sub-array under this condition.
Advanced:
If you have completed the O(n) time complexity solution, please try the O(n log n) time complexity solution.
Source: LeetCode
Link: https://leetcode-cn.com/problems/minimum-size-subarray-sum/
Solve
class Solution {
private:
int getSum(const vector<int> &nums, int i, int j) {
return std::accumulate(nums.begin() + i, nums.begin() + j + 1, 0);
}
public:
// 方法一,暴力解法,超时,时间复杂度O(N*N*N)
int minSubArrayLen_1e(int s, vector<int> &nums) {
const int n = nums.size();
int res = n + 1;
for (int i = 0; i < n; ++i) {
for (int j = i; j < n; ++j) {
if (getSum(nums, i, j) >= s) {
res = std::min(res, j - i + 1);
break; // 以i开始的连续子串求和已经大于0,后面的循环无效,跳出当前循环
}
}
}
return res == n + 1 ? 0 : res;
}
// 方法二,暴力解法,优化求和的计算,优化以i开始的子串满足条件后跳出循环,低效率通过,时间复杂度O(N*N)
int minSubArrayLen_2e(int s, vector<int> &nums) {
const int n = nums.size();
int res = n + 1;
for (int i = 0; i < n; ++i) {
int sum = 0;
for (int j = i; j < n; ++j) {
sum += nums[j]; // 这步优化将时间复杂度从O(N*N*N)提高到O(N*N)
if (sum >= s) {
res = std::min(res, j - i + 1);
break; // 以i开始的连续子串求和已经大于0,后面的循环无效,跳出当前循环
}
}
}
return res == n + 1 ? 0 : res;
}
// 方法三,双指针滑动窗口, O(N)
int minSubArrayLen_3e(int s, vector<int> &nums) {
const int n = nums.size();
int res = n + 1;
int l = 0;
int r = -1; // 循环不变量,[l...r]保存子串
int sum = 0;
while (l < n) {
if (sum < s && r + 1 < n) {
sum += nums[++r];
} else {
// r已经到底,即r == n-1
// 或者sum >= s
// 或者以上两者都满足
// ???如果 sum >= s,直接在这儿减掉一个nums[l]会不会导致有场景没有计入res
// 不会,出现上述情况不会走到else这个分支里面,会进行res计算,下次循环才跳入else中,和减去nums[l],且l递增
sum -= nums[l++];
}
if (sum >= s) {
res = std::min(res, r - l + 1);
}
}
return res == n + 1 ? 0 : res;
}
// 方法四,双指针滑动窗口另一种写法, O(N)
int minSubArrayLen_4e(int s, vector<int> &nums) {
const int n = nums.size();
int res = n + 1;
int l = 0;
int r = 0; // 循环不变量,[l...r] 保存子串
int sum = 0;
while (r < n) {
sum += nums[r];
while (sum >= s) {
// ???如果nums[r] >> s,则递进到此会不会出现l累加到超过r的情况
// 不会,如果出现上述情况,l同样递进到r的位置,和重新变为0,也就是在下次循环中l和r都从原来r + 1的位置重新开始计算循环
res = std::min(res, r - l + 1);
sum -= nums[l++];
}
++r;
}
return res == n + 1 ? 0 : res;
}
// 方法五,二分查找,O(N * LOGN)
int minSubArrayLen(int s, vector<int> &nums) {
const int n = nums.size();
vector<int> sum(n + 1, 0); // sum[i]表示nums[0....i-1]的和
for (int i = 1; i <= n; ++i) {
sum[i] = sum[i - 1] + nums[i - 1];
}
int res = n + 1;
for (int i = 1; i <= n; ++i) {
int tasget = sum[i - 1] + s;
auto iter = std::lower_bound(sum.cbegin(), sum.cend(), tasget);
if (iter != sum.cend()) {
res = std::min(res, static_cast<int>(std::distance(sum.cbegin(), iter)) - (i - 1));
}
}
return res == n + 1 ? 0 : res;
}
// 重写指针对撞方法
int minSubArrayLen_review1(int s, vector<int> &nums) {
const int n = nums.size();
int l = 0;
int r = 0; // [l....r]存储子串
int sum = 0;
int res = n + 1;
while (r < n) {
sum += nums[r];
while (sum >= s) {
res = std::min(res, r - l + 1);
sum -= nums[l++];
}
++r;
}
return res == n + 1 ? 0 : res;
}
};