Leetcode剑指Offer刷题 - 第四天

剑指 Offer 03. 数组中重复的数字

解法一：HashSet

class Solution {
    public int findRepeatNumber(int[] nums) {
        Set<Integer> set = new HashSet<>();
        for (int num : nums) {
            if (!set.add(num)) {
                return num;
            }
        }
        return -1;
    }
}

解法二：辅助数组（哈希桶思想）

class Solution {
    public int findRepeatNumber(int[] nums) {
        int[] tmp = new int[nums.length];
        for (int i = 0; i < nums.length; ++i) {
            tmp[nums[i]]++;
            if (tmp[nums[i]] == 2) {
                return nums[i];
            }
        }
        return -1;
    }
}

解法三：排序

class Solution {
    public int findRepeatNumber(int[] nums) {
        Arrays.sort(nums);
        for (int i = 1; i < nums.length; ++i) {
            if (nums[i - 1] == nums[i]) {
                return nums[i];
            }
        }
        return -1;
    }
}

解法四：原地交换

class Solution {
    public int findRepeatNumber4(int[] nums) {
        for (int i = 0; i < nums.length; ++i) {
            while (nums[i] != i) {
                //遇到重复元素
                if (nums[i] == nums[nums[i]]) {
                    return nums[i];
                }

                //交换
                int tmp = nums[i];
                nums[i] = nums[tmp];
                nums[tmp] = tmp;
            }
        }

        return -1;
    }
}

剑指 Offer 53 - I. 在排序数组中查找数字 I

解法一：暴力破解

class Solution {
    public int search(int[] nums, int target) {
        int count = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == target) {
                count++;
            }
        }
        
        return count;
    }
}

解法二：HashMap

class Solution {
    public int search(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        return map.get(target) == null ? 0 : map.get(target);

    }
}

解法三：二分法

class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (target < nums[mid]) {
                //比mid小，在左侧
                right = mid - 1;
            } else if (target > nums[mid]) {
                //比mid大，在右侧
                left = mid + 1;
            } else {
                //等于mid
                if (target != nums[left]) {
                    //left像右缩
                    left++;
                } else if (target != nums[right]) {
                    //right像左缩
                    right--;
                } else {
                    //说明就在这个区间内
                    break;
                }
            }
        }

        return right - left + 1;
    }
}

剑指 Offer 53 - II. 0～n-1中缺失的数字

解法一：遍历顺序

class Solution {
    public int missingNumber(int[] nums) {
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] != i) {
                return i;
            }
        }
        return nums.length;
    }
}

解法二：二分

class Solution {
    public int missingNumber(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (nums[mid] != mid) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}