int lengthOfLongestSubstring(string s) {

/* 排除第一种情况 */

if (0 == s.size())

return 0;

int startIdx = 0, maxLen = 1;

int currLen = 0;

int charTable[256]; // 这里去初始化一下这个

memset(charTable, -1, sizeof(int) * 256); // 做一个hash映射，用空间换时间， char->index

for (int i = 0; i < s.size(); i++) {

if (charTable[s[i]] != -1) { //判断是否在

int charIdx = charTable[s[i]];

if (charIdx >= startIdx) { // 说明有

maxLen = MAX(maxLen, currLen);

startIdx = charIdx + 1;

currLen = i - startIdx; // 如果确认有重复则当前位置，减去开始位置就是currLen

}

}

charTable[s[i]] = i;

currLen ++;

}

return MAX(maxLen, currLen);

}

Solution s;

assert(0 == s.lengthOfLongestSubstring("")); // 极端值覆盖

// 题目中的例子

assert(3 == s.lengthOfLongestSubstring("abcabcbb"));

assert(1 == s.lengthOfLongestSubstring("bbbbb"));

assert(3 == s.lengthOfLongestSubstring("pwwkew"));

assert(1 == s.lengthOfLongestSubstring(" "));

assert(3 == s.lengthOfLongestSubstring("dvdf"));

}