https://leetcode.com/problems/minimum-size-subarray-sum/#/solutions
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. For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint. click to show more practice. More practice: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
Since the given array contains only positive integers, the subarray sum can only increase by including more elements. Therefore, you don't have to include more elements once the current subarray already has a sum large enough. This gives the linear time complexity solution by maintaining a minimum window with a two indices.
窗口指针
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;
}
O(NlogN)
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;
}
为什么我这么做不对?
public int minSubArrayLen(int s, int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
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 = Arrays.binarySearch(sums, i + 1, sums.length, sums[i] + s);
if (end == sums.length) break;
if (end < 0) {
end = -(end + 1);
}
if (end - i < minLen) minLen = end - i;
}
return minLen == Integer.MAX_VALUE ? 0 : minLen;
}
Input:7 [2,3,1,2,4,3]
Output:1
Expected:2