android-notes/blogs/Algorithm/剑指 Offer/数组相关.md

数组相关

03. 数组中重复的数字

class Solution {
    public int findRepeatNumber(int[] nums) {
        int[] res = new int[nums.length];
        for (int i : nums) {
            res[i]++;
            if (res[i] > 1) return i;
        }
        return -1;
    }
}

class Solution {

    public int findRepeatNumber(int[] nums) {
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] == nums[i + 1]) {
                return nums[i];
            }
        }
        return -1;
    }
}

class Solution {

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

04. 二维数组中的查找

class Solution {

    public boolean findNumberIn2DArray(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        int m = matrix.length, n = matrix[0].length;
        int row = 0, col = n - 1;
        while (row < m && col >= 0) {
            if (matrix[row][col] > target) {
                col--;
            } else if (matrix[row][col] < target) {
                row++;
            } else {
                return true;
            }
        }
        return false;
    }
}

11. 旋转数组的最小数字

class Solution {

    public int minArray(int[] numbers) {
        int i = 0, j = numbers.length - 1;
        while (i < j) {
            int m = (i + j) / 2;
            if (numbers[m] > numbers[j]) {
                i = m + 1;
            } else if (numbers[m] < numbers[j]) {
                j = m;
            } else {
                j--;
            }
        }
        return numbers[i];
    }
}

21. 调整数组顺序使奇数位于偶数前面

class Solution {

    public int[] exchange(int[] nums) {
        int i = 0, j = nums.length - 1;
        while (i < j) {
            if (nums[i] % 2 == 1 && nums[j] % 2 == 0) {
                i++;
                j--;
            } else if (nums[i] % 2 == 1 && nums[j] % 2 == 1) {
                i++;
            } else if (nums[i] % 2 == 0 && nums[j] % 2 == 1) {
                int temp = nums[i];
                nums[i++] = nums[j];
                nums[j--] = temp;
            } else {
                j--;
            }
        }
        return nums;
    }
}

class Solution {

    public int[] exchange(int[] nums) {
        int i = 0, j = nums.length - 1, temp;
        while (i < j) {
            while (i < j && (nums[i] & 1) == 1) {
                i++;
            }
            while (i < j && (nums[j] & 1) == 0) {
                j--;
            }
            temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
        return nums;
    }
}

39. 数组中出现次数超过一半的数字

class Solution {

    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length / 2];
    }
}

class Solution {

    public int majorityElement(int[] nums) {
        int count = 0, result = 0;
        for (int num : nums) {
            if (count == 0) {
                result = num;
            }
            count += (result == num) ? 1 : -1;
        }
        return result;
    }
}

42. 连续子数组的最大和

class Solution {

    public int maxSubArray(int[] nums) {
        int max = Integer.MIN_VALUE;
        int sum = 0;
        for (int num : nums) {
            if (sum <= 0) {
                sum = num;
            } else {
                sum += num;
            }
            max = Math.max(max, sum);
        }
        return max;
    }
}

class Solution {

    public int maxSubArray(int[] nums) {
        int res = nums[0];
        for (int i = 1; i < nums.length; i++) {
            nums[i] += Math.max(nums[i - 1], 0);
            res = Math.max(res, nums[i]);
        }
        return res;
    }
}

45. 把数组排成最小的数

class Solution {
    public String minNumber(int[] nums) {
        String[] strings = getStringArray(nums);
        Arrays.sort(strings, new Cop());

        StringBuilder result = new StringBuilder();
        for (String s : strings) {
            result.append(s);
        }
        return result.toString();
    }

    private String[] getStringArray(int[] nums) {
        String[] strings = new String[nums.length];
        for (int i = 0; i < nums.length; i++) {
            strings[i] = String.valueOf(nums[i]);
        }
        return strings;
    }

    class Cop implements Comparator<String> {
        @Override
        public int compare(String o1, String o2) {
            String s1 = o1 + o2;
            String s2 = o2 + o1;
            return s1.compareTo(s2);
        }
    }
}

class Solution {
    public String minNumber(int[] nums) {
        String[] strings = new String[nums.length];
        for (int i = 0; i < nums.length; i++) {
            strings[i] = String.valueOf(nums[i]);
        }
        Arrays.sort(strings, (o1, o2) -> (o1 + o2).compareTo(o2 + o1));

        StringBuilder result = new StringBuilder();
        for (String s : strings) {
            result.append(s);
        }
        return result.toString();
    }
}

56 - I. 数组中数字出现的次数

class Solution {
    public int[] singleNumbers(int[] nums) {
        int sum = 0;
        for (int i : nums) {
            sum = sum ^ i;
        }
        int flag = sum & (-sum);
        int res = 0;
        for (int i : nums) {
            if ((flag & i) != 0) {
                res = res ^ i;
            }
        }
        return new int[]{res, sum ^ res};
    }
}

56 - II. 数组中数字出现的次数 II

class Solution {
    public int singleNumber(int[] nums) {
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 2; i += 3) {
            if (nums[i] != nums[i + 2]) {
                return nums[i];
            }
        }
        return nums[nums.length - 1];
    }
}

class Solution {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i : nums) {
            int count = map.getOrDefault(i, 0);
            map.put(i, count + 1);
        }
        for (Integer i : map.keySet()) {
            if (map.get(i) == 1) {
                return i;
            }
        }
        return -1;
    }
}

class Solution {
    public int singleNumber(int[] nums) {
        int[] count = new int[32];
        for (int i : nums) {
            for (int j = 0; j < 32; j++) {
                count[j] += i & 1;
                i = i >>> 1;
            }
        }
        int res = 0, flag = 3;
        for (int i = 0; i < 32; i++) {
            res += (1 << i) * (count[i] % flag);
        }
        return res;
    }
}