leetcode update

content/posts/leetcode.md

@@ -1351,3 +1351,115 @@ func findMaxLength(nums []int) int {
 ### 总结
 
 本题和前几天的题目在思想上有异曲同工之妙, 也是前缀和的一种应用, 如果把0和1的数量差作为前缀和, 那么本题解题思路可简单概括为找到前缀和相等的位置的最远距离.
+
+## day20 2024-03-17
+
+### 57. Insert Interval
+
+You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.
+
+Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).
+
+Return intervals after the insertion.
+
+Note that you don't need to modify intervals in-place. You can make a new array and return it.
+
+![03179SRnh9piq8gy](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/03179SRnh9piq8gy.png)
+
+### 题解
+
+遍历intervals, 将interval复制到结果数组中. 判断要插入的interval的start和end是否在当前interval中
+
+1. 在范围内则将新interval的start设置为当前interval的start. 使用相同的方式判断end的范围.
+2. 若start或end不在当前interval范围内且小于下一个interval的start, 则将start或者end设置为新interval的start或end. 此时新interval构造完成, 将后面的interval原样复制到result中即可.
+
+### 代码
+
+```go
+func insert(intervals [][]int, newInterval []int) [][]int {
+    result := [][]int{}
+    start := newInterval[0]
+    end := newInterval[1]
+    insertInterval := []int{}
+    flag := 0
+    for _, value := range intervals{
+        if flag == 2{
+            result = append(result, value)
+        }else if flag == 0{
+            if start < value[0]{
+                insertInterval = append(insertInterval, start)
+                flag++
+                if end < value[0]{
+                    insertInterval = append(insertInterval, end)
+                    flag++
+                    result = append(result, insertInterval)
+                    result = append(result, value)
+                }else if end >= value[0] && end <= value[1]{
+                    insertInterval = append(insertInterval, value[1])
+                    flag++
+                    result = append(result, insertInterval)
+                }
+            }else if start >= value[0] && start <= value[1]{
+                insertInterval = append(insertInterval, value[0])
+                flag++
+                if end >= value[0] && end <= value[1]{
+                    insertInterval = append(insertInterval, value[1])
+                    flag++
+                    result = append(result, insertInterval)
+                }
+            }else{
+                result = append(result, value)
+            }
+        }else{
+            if end < value[0]{
+                insertInterval = append(insertInterval, end)
+                flag++
+                result = append(result, insertInterval)
+                result = append(result, value)
+            }else if end >= value[0] && end <= value[1]{
+                insertInterval = append(insertInterval, value[1])
+                flag++
+                result = append(result, insertInterval)
+            }
+        }
+    }
+    if flag == 0{
+        result = append(result, []int{start, end})
+    }else if flag == 1{
+        insertInterval = append(insertInterval, end)
+        result = append(result, insertInterval)
+    }
+    return result
+}
+```
+
+### 总结
+
+这种题思路上并没有特别之处, 但是判断逻辑比较繁琐, 需要耐心思考边界情况, 查看他人题解, 发现用原始数组与需要插入的interval做比较要比使用interval和原始数组的start和end做比较思路简单清晰得多, 这里还是对判断条件的变与不变理解的不够透彻, 需要插入的interval的start和end是不变的, 不断遍历原始数组与不变的start和end做比较要比使用不变的start和end去和不断变化的interval的start和end做比较判断起来容易得多. 找到不变的条件对于思路清楚的解决问题至关重要. 给出示例代码
+
+```go
+func insert(intervals [][]int, newInterval []int) [][]int {
+
+    res := make([][]int, 0)
+
+    i := 0
+
+    for ; i < len(intervals) && intervals[i][1] < newInterval[0]; i++ {
+        res = append(res, intervals[i])
+    }
+
+    for ; i < len(intervals) && intervals[i][0] <= newInterval[1]; i++ {
+        newInterval[0] = min(intervals[i][0], newInterval[0])
+        newInterval[1] = max(intervals[i][1], newInterval[1])
+    }
+
+    res = append(res, newInterval)
+
+    for i < len(intervals) {
+        res = append(res, intervals[i])
+        i++
+    }
+
+    return res
+}
+```