【2021秋招_微软软开面经】2020.10.13 微软三面
1.螺旋矩阵
写一个计算数字第一次出现的位置的函数
function spiralMatrix(n){ let matrix = new Array(n).fill(0).map(ele => new Array(n).fill(0)); helper(n, matrix, 0, n*n); return matrix; function helper(n, matrix, level, count) { if (n % 2 == 0 && level * 2 == n) return; if (n % 2 != 0) { if (level * 2 + 1 == n) { matrix[level][level] = count; return; } } for(let i = level; i < n - level; i++) { matrix[level][i] = count--; } for(let i = level + 1; i < n - level; i++) { matrix[i][n-level-1] = count--; } for(let i = n - 2 - level; i >= level; i--) { matrix[n-level-1][i] = count--; } for(let i = n - 2 - level; i > level; i--) { matrix[i][level] = count--; } helper(n, matrix, ++level, count); } }2.有序递增数组,寻找某个数字出现的次数,要求最小时间复杂度
写一个计算数字第一次出现的位置的函数
function getFirstPos(arr, target, l, r) { let i = l, j = r, mid; while ( i <= j) { mid = parseInt(i + (j-i)/2); if (arr[mid] > target) { j = mid - 1; } else if (arr[mid] < target) { i = mid + 1; } else { if (mid > l && arr[mid-1] == target) { j = mid - 1; } else { return mid; } } } return -1; }let res = getFirstPos([1,2,2,3,4], 2, 0, 4); console.log(res);
function getLastPos(arr, target, l, r) { let i = l, j = r, mid; while ( i <= j ) { mid = parseInt(i + (j-i)/2); if (arr[mid] > target) { j = mid - 1; } else if (arr[mid] < target) { i = mid + 1; } else { if (mid + 1 <= r && arr[mid+1] == target) { i = mid + 1; } else { return mid; } } } return -1; }
完整版
function getNums(nums, target) { let l = nums.length; let left = getFirstPos(nums, target, 0, l-1); let right = getLastPos(nums, target, 0, l-1); if (left == -1 || right == -1) return 0; return right - left + 1; } function biSearch(arr, target, l, r) { let i = l, j = r, mid; while ( i <= j) { mid = parseInt(i + (j-i)/2); if (arr[mid] == target) { return mid; } else if (arr[mid] < target) { i = mid + 1; } else { j = mid - 1; } } return -1; } function getFirstPos(arr, target, l, r) { let index = biSearch(arr, target, l, r); if (index != -1 && index > 0 && arr[index-1] == target) { index = getFirstPos(arr, target, l, index - 1); } return index; } function getLastPos(arr, target, l, r) { let index = biSearch(arr, target, l, r); if (index != -1 && index < r - 1 && arr[index+1] == target) { index = getLastPos(arr, target, index+1, r); } return index; } let res = getNums([1,2,3,3,4,4,4,5], 4); console.log(res);3.给一个由0和1构成的数组,计算0和1个数相等的最长子数组,返回其长度
function getLIS(arr) { let l = arr.length; let sum = new Array(l).fill(0); sum[0] = arr[0]; for(let i = 1; i < l; i++) { sum[i] = (arr[i] == 1)? sum[i-1] + 1 : sum[i-1] - 1; } let res = 0, i = 0, j = l - 1, loop = 0; while (loop < l) { i = loop, j = l - 1; while (j > i && sum[j] != sum[i]) --j; if (j > i && sum[j] == sum[i]) { res = Math.max(j - i, res); break; } loop++; } return res; } let res = getLIS([1,1,1,0,0,1]); console.log(res);#微软##面经##校招##前端工程师#