LeetCode Day 9

LeetCode0017

• 给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
• 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

• 示例:
• 输入："23"
• 输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

/**
 * @param {string} digits
 * @return {string[]}
 */
var letterCombinations = function (digits) {
    let result = [];
    if (digits && digits.length > 0) {
        let letterMap = [
            "",//0
            "",//1
            "abc",//2
            "def",//3
            "ghi",//4
            "jkl",//5
            "mno",//6
            "pqrs",//7
            "tuv",//8
            "wxyz"//9
        ]
        for (let c of digits) {
            let digit = c - '0';
            if (digit === 0 || digit === 1) continue;
            let letter = letterMap[digit];
            if (result.length === 0) result = letter.split('');
            else {
                let arr = result.concat();
                result = [];
                for (let l of letter) {
                    for (let i = 0, lens = arr.length; i < lens; i++) {
                        result.push(arr[i] + l);
                    }

                }
            }
        }
    }
    return result;
};

console.log(letterCombinations("23"));

LeetCode0018

• 给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。

• 例如：给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。

• 满足要求的四元组集合为：[ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]

思路

• 套用求三数和的思路，对于nums[0...n-3]，求出一个threeSum(nums[1...n-1], target-nums[0...n-3])的值。

• 剩下的写法就跟求threeSum一致了，需要注意的是每个位置的去重处理。

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[][]}
 */
var fourSum = function (nums, target) {
    if (nums === undefined) return [];
    let lens = nums.length;
    if (lens < 4) return [];
    let first = 0, second, head, tail;
    let result = [];
    nums.sort((a, b) => a - b);
    //console.log(nums);
    while (first < lens - 3) {
        second = first + 1;
        while (second < lens - 2) {
            head = second + 1;
            tail = lens - 1;
            while (head < tail) {
                let sum = nums[first] + nums[second] + nums[head] + nums[tail];
                if (sum === target) {
                    result.push([nums[first], nums[second], nums[head], nums[tail]]);
                    while (nums[head] === nums[head + 1])++head;
                    while (nums[tail] === nums[tail - 1])--tail;
                    head++;
                    tail--;
                } else if (sum > target) {
                    tail--;
                }
                else {
                    head++;
                }
            }
            second++;
            while (nums[second] === nums[second - 1]) second++;
        }
        first++;
        while (nums[first] === nums[first - 1]) first++;
    }
    return result;
};

console.log(fourSum([1, 0, -1, 0, -2, 2], 0));
// console.log(fourSum([-3, -2, -1, 0, 0, 1, 2, 3], 0));
// console.log(fourSum([1, 0, -1, 0, -2, 2], 0));