2020-07-31

Author: JiayangWu

0001.两数之和/0001-两数之和.js

@@ -0,0 +1,12 @@
+const twoSum = (nums, target) => {
+  const prevNums = {};                         // 存放出现过的数字，和对应的索引
+  for (let i = 0; i < nums.length; i++) {      // 遍历每一项
+    const curNum = nums[i];                    // 当前项
+    const targetNum = target - curNum;         // 希望从过去的数字中找到的呼应项
+    const targetNumIndex = prevNums[targetNum];// 在prevNums中找targetNum的索引
+    if (targetNumIndex !== undefined) {        // 如果能找到
+      return [targetNumIndex, i];              // 直接返回targetNumIndex和当前的i
+    }                                          // 如果找不到，说明之前没出现过targetNum
+    prevNums[curNum] = i;                      // 往prevNums存当前curNum和对应的i
+  }
+}

1529.灯泡开关IV/1529-灯泡开关IV.py

@@ -0,0 +1,19 @@
+class Solution(object):
+    def minFlips(self, target):
+        """
+        :type target: str
+        :rtype: int
+        """
+        i = 0
+        flag = 0 
+        pre = None
+        res = 0
+        for i, ch in enumerate(target):
+            if not flag and ch == "0":
+                continue 
+            else:
+                if ch != pre:
+                    res += 1
+                flag = 1
+                pre = ch 
+        return res

面试题02.06.回文链表/面试题02.06-回文链表.py

@@ -0,0 +1,18 @@
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+class Solution(object):
+    def isPalindrome(self, head):
+        """
+        :type head: ListNode
+        :rtype: bool
+        """
+        s = []
+        while head:
+            s.append(head.val)
+            head = head.next 
+        return s == s[::-1]
+            
+