to do

12.3] Best Time to Buy and Sell Stock

    int maxProfit(vector<int>& prices) {
        if (prices.size()<2) return 0;
        
        int lmin = prices[0];
        vector<int> profits(prices.size(), -1);
        for (int i=0; i<prices.size(); ++i) {
            profits[i] = prices[i] - lmin;
            lmin = min(lmin, prices[i]);
        }
        return *(max_element(profits.begin(), profits.end()));
    }

12.4] Best Time to Buy and Sell Stock II

    int maxProfit(vector<int>& prices) {
        int sum = 0;
        for (int i=1; i<prices.size(); ++i) {
            int prof = prices[i] - prices[i-1];
            if (prof>0) sum += prof;
        }
        return sum;
    }

3. Longest Substring Without Repeating Characters

这样会超时，还是hash记录。但是考虑记录什么才不用重复擦写

    // find index of first occurrence of chart c in s[l~r], -1 if no such duplicate
    int findDupIndex (string& s, int l, int r, char c) {
        for (int i=l; i<=r; ++i) {
            if (s[i] == c) return i;
        }
        return -1;
    }
    
    int lengthOfLongestSubstring(string s) {
        int maxLen = 0, starti = 0; 
        for (int i=0; i<s.size(); ) {
            int dupIndex = findDupIndex(s, starti, i-1, s[i]);
            if (dupIndex != -1) {
                maxLen = max(maxLen, i-starti);
                starti = dupIndex+1;
                i = starti;
            } else {
                ++i;
            }
        }
        return max(maxLen, (int)s.size()-starti);
    }

12.5] Longest Substring Without Repeating Characters

    int lengthOfLongestSubstring(string s) {
        unordered_map<char, int> lastOccurred;
        int starti=0, maxlen = 0;
        for (int i=0; i<s.size(); ++i) {
            if (lastOccurred.find(s[i]) != lastOccurred.end() && 
                lastOccurred[s[i]] >= starti) {
                maxlen = max(maxlen, i-starti);
                starti = lastOccurred[s[i]] + 1;
            } 
            lastOccurred[s[i]] = i;
        }
        return max(maxlen, (int)s.size()-starti);
    }

12.6] Container With Most Water

    int maxArea(vector<int>& height) {
        if (height.size()<2) return 0;
        int maxA = 0;
        for (int start=0, end=height.size()-1; start<=end; ) {
            maxA = max(maxA, min(height[start], height[end]) * (end-start));  
            if (height[start] < height[end]) {
                ++start;
            } else {
                --end;
            }
        }
        return maxA;
    }