O(nlogn)解法

lc209. Minimum Size Subarray Sum.py

@@ -0,0 +1,44 @@
+class Solution:
+    def minSubArrayLen(self, s, nums):
+        """
+        :type s: int
+        :type nums: List[int]
+        :rtype: int
+        """
+        def bin_search(left, right, key, sums):
+            while left <= right:
+                mid = (left + right) // 2
+                if sums[mid] >= key:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+            return left
+
+        length = len(nums)
+        sums = [None] * (length + 1)
+        sums[0] = 0
+
+        for i in range(0, length):
+            sums[i + 1] = sums[i] + nums[i]
+
+        # 两层循环 O(N^2)解法
+        # for i in range(1, length + 1):
+        #     for j in range(length + 1 - i):
+        #         if sums[j + i] - sums[j] >= s:
+        #             return i
+
+        res = length + 1
+        for i in range(length + 1):
+            # sums[right]-sums[left]，不包括nums[left]
+            tem = bin_search(i + 1, length, sums[i] + s, sums)
+            if tem == length + 1:
+                break
+            if tem - i < res:
+                res = tem - i
+
+        return 0 if res == length + 1 else res
+
+s=100
+nums=[]
+soln = Solution()
+print(soln.minSubArrayLen(s, nums))