- 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"));
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));