[LeetCode] 159. Longest Substring with At Most Two Distinct Characters

Given a string s , find the length of the longest substring t  that contains at most 2 distinct characters.

Example 1:

Input: "eceba"
Output: 3
Explanation: tis "ece" which its length is 3.

Example 2:

Input: "ccaabbb"
Output: 5
Explanation: tis "aabbb" which its length is 5.

最多有两个不同字符的最长子串。题意是给一个字符串，请返回一个最长子串的长度。子串的要求是最多只有两个不同的字母。这题依然是用到sliding window的思想。给左右两个指针，并且创建一个hashmap，key是遍历到的字母，value是字母最后一次出现位置的index。扫描的时候，一开始只让右指针移动，把遍历到的字母和当前的坐标值加入hashmap。此时最大子串的长度就是左右指针的间距。当右指针遍历到第三个不同字母的时候，需要扫描hashmap，找到value（记为leftmost）最小的那个key并且删除这个键值对。此时左指针的位置是leftmost + 1。

时间O(n)

空间O(1) - 虽然有hashmap但是size很小，只有26个字母

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstringTwoDistinct = function (s) {
    let map = new Map();
    let left = 0;
    let max = 0;
    for (let right = 0; right < s.length; right++) {
        map.set(s[right], right);
        if (map.size > 2) {
            let leftMostChar = null;
            let leftMostIndex = Infinity;
            for (let [char, index] of map) {
                if (index < leftMostIndex) {
                    leftMostChar = char;
                    leftMostIndex = index;
                }
            }
            map.delete(leftMostChar);
            left = leftMostIndex + 1;
        }
        max = Math.max(max, right - left + 1);
    }
    return max;
};

另外一种模板写法，其他同类题型都可用。

Java实现

class Solution {
    public int lengthOfLongestSubstringTwoDistinct(String s) {
        // corner case
        if (s == null || s.length() == 0) {
            return 0;
        }

        // normal case
        int[] map = new int[256];
        int start = 0;
        int end = 0;
        int maxLen = Integer.MIN_VALUE;
        int counter = 0;
        while (end < s.length()) {
            char c1 = s.charAt(end);
            if (map[c1] == 0) {
                counter++;
            }
            map[c1]++;
            end++;
            while (counter > 2) {
                char c2 = s.charAt(start);
                if (map[c2] == 1) {
                    counter--;
                }
                map[c2]--;
                start++;
            }
            maxLen = Math.max(maxLen, end - start);
        }
        return maxLen;
    }
}

JavaScript实现

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstringTwoDistinct = function(s) {
    // corner case
    if (s == null || s.length == 0) {
        return 0;
    }

    // normal case
    let start = 0;
    let end = 0;
    let maxLen = -Infinity;
    let counter = 0;
    let map = {};
    while (end < s.length) {
        let c1 = s.charAt(end);
        if (map[c1] == null || map[c1] <= 0) {
            counter++;
        }
        map[c1] = map[c1] + 1 || 1;
        end++;
        while (counter > 2) {
            let c2 = s.charAt(start);
            if (map[c2] == 1) {
                counter--;
            }
            map[c2]--;
            start++;
        }
        maxLen = Math.max(maxLen, end - start);
    }
    return maxLen;
};

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