Leetcode

Review:

977. Squares of a Sorted Array

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int n = nums.size();
        vector<int> res;
        int left = 0, right = n-1;
        while(left<=right){
            if(abs(nums[left]) < abs(nums[right])){
                res.push_back(nums[right]*nums[right]);
                right--;
            }else{
                res.push_back(nums[left]*nums[left]);
                left++;
            }
        }
        reverse(res.begin(), res.end());
        return res;

    }
};

滑动窗口：

209. Minimum Size Subarray Sum

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int left = 0, right = 0;
        int sum = 0;
        int minVal = INT_MAX;

        while(right < nums.size()){
            sum += nums[right];
            right++;
            while(sum >= target && left < right){
                minVal = min(minVal, right-left);
                sum -= nums[left];
                left++;
            }
        }

        return minVal == INT_MAX ? 0 : minVal;

    }
};

3. Longest Substring Without Repeating Characters（*）

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int left = 0, right = 0;
        unordered_map<int, int> map;
        int res = 0;

        while(right < s.size()){
            char ch = s[right];
            map[ch]++;
            right++;
            while(map[ch] > 1){
                map[s[left]] --;
                left++;
            }
            res = max(res, right-left);
        }
        return res;
    }
};

自己写出现的问题：

这里我有一个地方想错了，重复的两个未必是两端，这样的话，就要把重复之前的全部都删掉，所以缩小窗口的时候，不能是map[s[right]]--  (只删掉重复的，而是s[left], 这样就能把重复之前的数字全部都删掉)

New:

125. Valid Palindrome

class Solution {
public:
    bool isPalindrome(string s) {
        string convStr;
        for(int i=0; i<s.size(); i++){
            if(isalnum(s[i])){
                convStr += tolower(s[i]);
            }
        }

        int left = 0, right = convStr.size()-1;
        while(left <= right){
            if(convStr[left] != convStr[right])
                return false;
            left++;
            right--;
        }

        return true;
    }
};

两个函数

isalnum

tolower