From 5c5f32bd0a77591f41d28a524b53e63c384a2e5c Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Thu, 14 Mar 2024 11:30:50 +0800
Subject: [PATCH] leetcode update

---
 content/posts/leetcode.md | 174 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 174 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 26aa282..46c7ed3 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -1042,3 +1042,177 @@ func pivotInteger(n int) int {
 ### 总结
 
 查看他人解法发现本题其实也可以使用前缀和进行求解, 将每个下标位置处的前缀和求出并保存, 倒序遍历数组, 将后面的数的和与当前位置的前缀和比较即可, 示例代码
+
+## day17 2024-03-14
+
+### 930. Binary Subarrays With Sum
+
+Given a binary array nums and an integer goal, return the number of non-empty subarrays with a sum goal.
+
+A subarray is a contiguous part of the array.
+
+Example 1:
+
+Input: nums = [1,0,1,0,1], goal = 2
+Output: 4
+Explanation: The 4 subarrays are bolded and underlined below:
+[1,0,1,0,1]
+[1,0,1,0,1]
+[1,0,1,0,1]
+[1,0,1,0,1]
+Example 2:
+
+Input: nums = [0,0,0,0,0], goal = 0
+Output: 15
+
+### 题解
+
+本题可以用前缀和求解, 类似这种求数组中连续和的问题都可以用前缀和求解, 首先计算出所有位置上的前缀和, 使用滑动窗口遍历前缀和数组, 根据goal的值调整窗口的左右边界, 注意0需要特殊处理, 0是否存在不影响整个序列的和, 0所在的位置处的前缀和和前面的元素相同.
+
+### 代码
+
+```go
+func numSubarraysWithSum(nums []int, goal int) int {
+
+    sum := 0
+
+    prefixsum := []int{0}
+
+    for _,value := range nums{
+
+        sum += value
+
+        prefixsum = append(prefixsum, sum)
+
+    }
+
+    front := 1
+
+    back := 0
+
+    flag := 0
+
+    result := 0
+
+    for front < len(prefixsum){
+
+        if prefixsum[front] - prefixsum[back]<goal{
+
+            front++
+
+        }else if prefixsum[front] - prefixsum[back]>goal{
+
+            if front-1 == back{
+
+                front++
+
+            }
+
+            back++
+
+        }else{
+
+            if (front-1 == back){
+
+                result++
+
+                front++
+
+                back = flag
+
+            }else if prefixsum[back] == prefixsum[back+1]{
+
+                result++
+
+                back++
+
+            }else if (front<len(prefixsum)-1) && prefixsum[front] == prefixsum[front+1]{
+
+                result++
+
+                front++
+
+                back = flag
+
+            }else{
+
+                result++
+
+                back++
+
+                flag=back
+
+            }
+
+        }
+
+    }
+
+    return result
+
+}
+```
+
+### 总结
+
+本题使用滑动窗口时间复杂度上比较高, 考虑在滑动窗口的过程中, 连续的0的部分被重复遍历, 大大增加了总体的运行时间. 其实本题只需要用当前的前缀和减去goal, 并寻找前面的前缀和中是否有符合差值的前缀和存在, 存在则从该前缀和位置到当前前缀和位置直接的序列满足和为goal的条件. 已经求出了前缀和就没必要再去一个个遍历并且滑动了, 如果使用滑动窗口则没必要计算前缀和. 我的解法实际上是对前缀和理解不深刻导致的. 巧用前缀和加哈希表(存储某个前缀和出现的次数)可以快速的解决这个问题. 示例代码如下
+
+```go
+func numSubarraysWithSum(nums []int, goal int) int {
+    hash := map[int]int{}
+    sum := 0
+    count := 0
+    hash[0] = 1
+    for i:=0; i < len(nums); i++ {
+        sum = sum + nums[i]
+        count = count + hash[sum - goal]
+        val, ok := hash[sum]
+        if(ok) {
+            hash[sum] = val + 1
+        } else {
+            hash[sum] = 1
+        }
+    }
+    return count
+}
+```
+
+还有一种极其巧妙的解法, 分别求得和小于等于goal的连续子序列的数量, 再减去和小于等于goal-1的连续子序列的数量即为最终结果. 本题中求子序列等于goal是比较困难的, 要考虑很多条件, 但是求小于等于goal的子序列却是比较简单的问题. 这种方法将一个困难问题转化为两个简单的子问题求解, 得到了更高效的方法. 充分利用整体性可使问题更简单. 代码如下
+
+```go
+func numSubarraysWithSum(nums []int, goal int) int {
+
+	return counting(nums, goal) - counting(nums, goal-1)
+
+}
+
+func counting(nums []int, goal int) int {
+
+	if goal < 0 {
+
+		return 0
+
+	}
+
+	left, right, sum, ans := 0, 0, 0, 0
+
+	for ; right < len(nums); right++ {
+
+		sum += nums[right]
+
+		for left <= right && sum > goal {
+
+			sum -= nums[left]
+
+			left++
+
+		}
+
+		ans += right - left + 1
+
+	}
+
+	return ans
+
+}
+```