用时：42min
class Solution {public int removeDuplicates(int[] nums) {/*** 双指针，右指针遍历整个数组，左指针记录有效值*/int l = 0, r = 0;Set<Integer> s = new HashSet<Integer>();for(; r < nums.length; r++){if(s.add(nums[r])){int t = nums[l];nums[l] = nums[r];nums[r] = t;l++;}}return l;}
}/*优化后版本*/
class Solution {public int removeDuplicates(int[] nums) {// 非严格递增序列：去除重复元素后的序列是递增的，也就是说重复元素相邻int l = 0, r = 0;for(; r < nums.length; r++){// 让左指针位置的元素是符合条件的，右指针一直向后遍历// 换句话说就是让左指针后面的相邻元素不与左指针元素重复if(nums[l] != nums[r]){nums[l + 1] = nums[r];l++;}}return l + 1;}
}
用时: 17min
class Solution {public int maxProfit(int[] prices) {// 当天买卖无收益，计算相邻两天的正收益之和 [贪心算法：通过局部最优解实现整体最优]// 假如是第一天买，第三天卖 (d3 - d1) = (d3 - d2) + (d2 - d1)int profit = 0;for(int i = 1; i < prices.length; i++){// profit += prices[i] - prices[i - 1] > 0 ? prices[i] - prices[i - 1] : 0;profit += Math.max(0, prices[i] - prices[i - 1]);}return profit;}
}
用时：38min
class Solution {public void rotate(int[] nums, int k) {/**先整体翻转，再局部翻转*/k %= nums.length;convertNums(nums, 0, nums.length - 1);convertNums(nums, 0, k - 1);convertNums(nums, k, nums.length - 1);}public void convertNums(int [] nums, int l , int r){while(l < r){int t = nums[l];nums[l] = nums[r];nums[r] = t;l++;r--;}}
}
用时：5min
class Solution {public boolean containsDuplicate(int[] nums) {// 占用额外内存Set<Integer> s = new HashSet<>();for(int num : nums){if(!s.add(num)){return true;}}return false;// 慢// Arrays.sort(nums);// for(int i = 1; i < nums.length; i++){// if(nums[i - 1] == nums[i]){// return true;// }// }// return false;}
}
用时: 26min
class Solution {public int singleNumber(int[] nums) {if(nums.length == 1){return nums[0];}Arrays.sort(nums);int t = nums[0];int p = -1;for(int i = 1; i < nums.length; i++){if(nums[i] == p){continue;}if(t == nums[i] && t != p){p = nums[i];t = -1;}else{t = t == -1 ? nums[i] : t;}}return t;}
}/* 优化版本 */
class Solution {public int singleNumber(int[] nums) {// 位运算（异或运算，相同为零、不同为1, 0 ^ n = n，所以最后就剩下那个唯一的元素）if(nums.length == 1){return nums[0];}int result = 0;for(int num : nums){result ^= num;}return result;}
}
用时: 32min
class Solution {public int[] intersect(int[] nums1, int[] nums2) {// 散列桶计数Map<Integer, Integer> map = new HashMap<Integer, Integer>();for(int num : nums1){map.put(num, map.getOrDefault(num, 0) + 1);}int [] result = new int[nums1.length < nums2.length ? nums1.length : nums2.length];int index = -1;for(int num : nums2){int count = map.getOrDefault(num, 0);if(count > 0){result[++index] = num;map.put(num, count - 1);}}return Arrays.copyOfRange(result, 0, index + 1);}
}
用时: 33min
class Solution {public int[] plusOne(int[] digits) {int len = digits.length;boolean last = false;for(int i = len - 1; i >= 0; i--){int num = digits[i] + 1;if(num < 10){digits[i] = num;break;}else{digits[i] = num - 10;}if(i == 0){last = num > 9;}}int [] arr = last ? new int[digits.length + 1] : new int[digits.length];if(last){arr[0] = 1;System.arraycopy(digits, 0, arr, 1, digits.length);return arr;}else{return digits;}}
}/* 优化版本 */
class Solution {public int[] plusOne(int[] digits) {for(int i = digits.length - 1; i >= 0; i--){int num = digits[i] + 1;digits[i] = num % 10;if(num < 10){return digits;}}digits = new int[digits.length + 1];// 为什么之赋值一个就够了呢？因为原数组空间不够的情况只能是数值为 10000(n个零)digits[0] = 1; return digits;}
}
用时：5min
class Solution {public void moveZeroes(int[] nums) {int l = 0, r = 0;while(r < nums.length){if(nums[r] != 0){int t = nums[l];nums[l] = nums[r];nums[r] = t;l++;}r++;}}
}
用时：9min
class Solution {public int[] twoSum(int[] nums, int target) {Map<Integer, Integer> map = new HashMap<>();for(int i = 0; i < nums.length; i++){int num = target - nums[i];int idx = map.getOrDefault(num, -1);if(idx != -1){return new int[]{idx, i};}map.put(nums[i], i);}return new int[]{-1, -1};}
}
用时：20min
class Solution {public boolean isValidSudoku(char[][] board) {int [][] row = new int[9][9];int [][] column = new int[9][9];int [][][] block = new int[3][3][9];for(int i = 0; i < board.length; i++){for(int j = 0; j < board[0].length; j++){if(board[i][j] == '.'){continue;}int idx = board[i][j] - '0' - 1; // 数值要转化成对应索引，所以需要减一row[i][idx]++;column[j][idx]++;block[i / 3][j / 3][idx]++;if(row[i][idx] > 1 || column[j][idx] > 1 || block[i / 3][j / 3][idx] > 1){return false; }}}return true;}
}
用时：15min
class Solution {public void rotate(int[][] matrix) {/** 先主对角线翻转，然后再左右翻转 */int [][] arr = new int[matrix.length][matrix[0].length];for(int i = 0; i < matrix.length; i++){for(int j = 0; j < i; j++){int num = matrix[i][j];matrix[i][j] = matrix[j][i];matrix[j][i] = num;}}for(int i = 0; i < matrix.length; i++){for(int j = 0; j < matrix[0].length / 2; j++){int num = matrix[i][j];matrix[i][j] = matrix[i][matrix[0].length -1 - j];matrix[i][matrix[0].length -1 - j] = num;}}}
}
