這里介紹排序演算法及貪心演算法的個人解決方法
題目一：兩數之和
給定一個整數陣列 nums 和一個整數目標值 target，請你在該陣列中找出 和為目標值 的那 兩個 整數，并回傳它們的陣列下標， 你可以假設每種輸入只會對應一個答案，但是，陣列中同一個元素不能使用兩遍， 你可以按任意順序回傳答案，
示例 1：
輸入：nums = [2,7,11,15], target = 9 輸出：[0,1] 解釋：因為 nums[0] + nums[1] == 9 ，回傳 [0, 1] ，
示例 2：
輸入：nums = [3,2,4], target = 6 輸出：[1,2]
示例 3：
輸入：nums = [3,3], target = 6 輸出：[0,1]
提示：
2 <= nums.length <= 103 -109 <= nums[i] <= 109 -109 <= target <= 109 只會存在一個有效答案
答案：
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
let arr=[];
for(let i=0;i<nums.length-1;i++){
let bind = true;
let jj = 0;
for(let j=i+1;j<nums.length;j++){
if(nums[i]+nums[j] == target){
bind = false;
jj = j;
}
}
if(!bind){
arr.push(i)
arr.push(jj)
nums.splice(i,1);
nums.splice(jj,1);
i--;
}
}
return arr;
};
題目二：有效的數獨
判斷一個 9x9 的數獨是否有效，只需要根據以下規則，驗證已經填入的數字是否有效即可， 1.數字 1-9 在每一行只能出現一次， 2.數字 1-9 在每一列只能出現一次， 3.數字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現一次，
上圖是一個部分填充的有效的數獨， 數獨部分空格內已填入了數字，空白格用 '.' 表示，
示例 1:
輸入: [ ["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 輸出: true
示例 2:
輸入: [ ["8","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 輸出: false
說明:
一個有效的數獨（部分已被填充）不一定是可解的， 只需要根據以上規則，驗證已經填入的數字是否有效即可， 給定數獨序列只包含數字 1-9 和字符 '.' ， 給定數獨永遠是 9x9 形式的，
答案：
/**
* @param {character[][]} board
* @return {boolean}
*/
var isValidSudoku = function(board) {
for (let arr of board) {
let row = []
for (let c of arr) {
if (c !== '.') row.push(c);
}
let set = new Set(row)
if (set.size !== row.length) return false;
}
// 檢查每一列
for (let i = 0; i < 9; i++) {
let col = []
board.map( arr => {
if (arr[i] !== '.') col.push(arr[i])
})
let set = new Set(col)
if (set.size !== col.length) return false;
}
// 檢查每個小方塊
for (let x = 0; x < 9; x += 3) {
for (let y = 0; y < 9; y += 3) {
let box = []
for (let a = x; a < 3 + x; a ++) {
for (let b = y; b < 3 + y; b ++) {
if (board[a][b] !== '.') box.push(board[a][b])
}
}
let set = new Set(box)
if (set.size !== box.length) return false
}
}
return true
};
題目三：旋轉影像
給定一個 n × n 的二維矩陣 matrix 表示一個影像，請你將影像順時針旋轉 90 度， 你必須在 原地 旋轉影像，這意味著你需要直接修改輸入的二維矩陣，請不要 使用另一個矩陣來旋轉影像，
示例 1：
輸入：matrix = [[1,2,3],[4,5,6],[7,8,9]] 輸出：[[7,4,1],[8,5,2],[9,6,3]]
示例 2：
輸入：matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] 輸出：[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
示例 3：
輸入：matrix = [[1]] 輸出：[[1]]
示例 4：
輸入：matrix = [[1,2],[3,4]] 輸出：[[3,1],[4,2]]
提示：
matrix.length == n matrix[i].length == n 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000
答案：
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var rotate = function(matrix) {
let length = matrix.length;
for(let i=0;i<length/2;i++){
for (let j = i; j < length - 1 - i; j++) {
let tmp = matrix[i][j];
matrix[i][j] = matrix[length - 1 - j][i];
matrix[length - 1 - j][i] = matrix[length - 1 - i][length - 1 - j];
matrix[length - 1 - i][length - 1 - j] = matrix[j][length - 1 - i];
matrix[j][length - 1 - i] = tmp;
}
}
return matrix;
};
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