【LBLD】双指针技巧秒杀七道数组题目
快慢指针技巧
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int fast = 0;
int slow = 0;
while (fast < nums.size()) {
if (nums.at(fast) != nums.at(slow)) {
nums.at(++slow) = nums.at(fast);
}
fast++;
}
return slow+1;
}
};
左右指针的常用算法
2、两数之和
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int left = 0, right = numbers.size()-1;
int sum = numbers[left] + numbers[right];
while (left < right) {
if (sum == target) {
return {left+1, right+1};
}
else if (sum < target) {
left++;
}
else {
right--;
}
sum = numbers[left] + numbers[right];
}
return {-1, -1};
}
};
3、反转数组
class Solution {
public:
void reverseString(vector<char>& s) {
int left = 0, right = s.size()-1;
char temp = '\0';
while (left < right) {
temp = s[left];
s[left] = s[right];
s[right] = temp;
left++; right--;
}
}
};
4、回文串判断
class Solution {
public:
string palindromeLength(string s, int i, int j) {
int left = i, right = j;
while (left >= 0 && right < s.size() && s[left] == s[right]) {
left--; right++;
}
return s.substr(left+1, right-left-1);
}
string longestPalindrome(string s) {
string ans = "";
string res1 = "", res2 = "";
for (int i = 0; i < s.size(); i++) {
res1 = palindromeLength(s, i, i);
res2 = palindromeLength(s, i, i+1);
ans = ans.size() > res1.size() ? ans : res1;
ans = ans.size() > res2.size() ? ans : res2;
}
return ans;
}
};
标签:size,right,int,numbers,ans,七道,LBLD,指针,left
From: https://www.cnblogs.com/yangxuanzhi/p/17275633.html