leetcode 235-290 easy
235. Lowest Common Ancestor of a Binary Search Tree
公共的祖先必定大于左点小于右点,否则不断递归到合适。
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if ((root -> val > p -> val) && (root -> val > q -> val)) {
return lowestCommonAncestor(root -> left, p, q);
}
if ((root -> val < p -> val) && (root -> val < q -> val)) {
return lowestCommonAncestor(root -> right, p, q);
}
return root;
}
}; ///////////////////////////////// class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* cur = root;
while (true) {
if ((cur -> val > p -> val) && (cur -> val > q -> val)) {
cur = cur -> left;
} else if ((cur -> val < p -> val) && (cur -> val < q -> val)) {
cur = cur -> right;
} else {
return cur;
}
}
}
};
257. Binary Tree Paths
void binaryTreePaths(vector<string>& result, TreeNode* root, string t) {
if(!root->left && !root->right) {
result.push_back(t);
return;
} if(root->left) binaryTreePaths(result, root->left, t + "->" + to_string(root->left->val));
if(root->right) binaryTreePaths(result, root->right, t + "->" + to_string(root->right->val));
} vector<string> binaryTreePaths(TreeNode* root) {
vector<string> result;
if(!root) return result; binaryTreePaths(result, root, to_string(root->val));
return result;
}
258. Add Digits
Iteration method class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
while(num >= 10):
temp = 0
while(num > 0):
temp += num % 10
num /= 10
num = temp
return num
Digital Root this method depends on the truth: N=(a[0] * 1 + a[1] * 10 + ...a[n] * 10 ^n),and a[0]...a[n] are all between [0,9] we set M = a[0] + a[1] + ..a[n] and another truth is that: 1 % 9 = 1 10 % 9 = 1 100 % 9 = 1 so N % 9 = a[0] + a[1] + ..a[n] means N % 9 = M so N = M (% 9) as 9 % 9 = 0,so we can make (n - 1) % 9 + 1 to help us solve the problem when n is 9.as N is 9, ( 9 - 1) % 9 + 1 = 9
///就是一个数的数根等于该数各位数的和的mod 9
/// (num-1)%9+1 等于 num%9,这为了解决9的树根时9而不是0的问题
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
if num == 0 : return 0
else:return (num - 1) % 9 + 1
263. Ugly Number
bool isUgly(int num) {
if(num == ) return false; while(num% == ) num/=;
while(num% == ) num/=;
while(num% == ) num/=; return num == ;
}
268. Missing Number
1.XOR
相同则为0。num[]的数字和下标一样
public int missingNumber(int[] nums) { //xor
int res = nums.length;
for(int i=0; i<nums.length; i++){
res ^= i;
res ^= nums[i];
}
return res;
}
2.SUM
public int missingNumber(int[] nums) { //sum
int len = nums.length;
int sum = (0+len)*(len+1)/2;
for(int i=0; i<len; i++)
sum-=nums[i];
return sum;
}
3.Binary Search
public int missingNumber(int[] nums) { //binary search
Arrays.sort(nums);
int left = 0, right = nums.length, mid= (left + right)/2;
while(left<right){
mid = (left + right)/2;
if(nums[mid]>mid) right = mid;
else left = mid+1;
}
return left;
}
Summary:
If the array is in order, I prefer Binary Search method. Otherwise, the XOR method is better.
283. Move Zeroes
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int j = ;
// move all the nonzero elements advance
for (int i = ; i < nums.size(); i++) {
if (nums[i] != ) {
nums[j++] = nums[i];
}
}
for (;j < nums.size(); j++) {
nums[j] = ;
}
}
};
290. Word Pattern
Input: pattern ="abba"
, str ="dog cat cat dog"
Output: true
bool wordPattern(string pattern, string str) {
map<char, int> p2i;
map<string, int> w2i;
istringstream in(str);
int i = , n = pattern.size();
for (string word; in >> word; ++i) {
if (i == n || p2i[pattern[i]] != w2i[word])
return false;
p2i[pattern[i]] = w2i[word] = i + ;
}
return i == n;
}
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