上岸算法 I LeetCode Weekly Contest 227解题报告

No.1 检查数组是否经排序和轮转得到

解题思路

枚举、检查。

代码展示

class Solution {
    public boolean check(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (check(nums, i)) {
                return true;
            }
        }
        return false;
    }

    private boolean check(int[] nums, int i) {
        for (int j = 1; j < nums.length; j++) {
            if (nums[(i + j - 1 + nums.length) % nums.length] > nums[(i + j) % nums.length]) {
                return false;
            }
        }
        return true;
    }
}

No.2 移除石子的最大得分

解题思路

贪心，每次移除最多的和最少的。

代码展示

class Solution {
    public int maximumScore(int a, int b, int c) {
        int res = 0;
        int[] nums = {a, b, c};
        Arrays.sort(nums);
        for (; nums[1] != 0; res++) {
            if (nums[0] == 0) {
                nums[1]--;
            } else {
                nums[0]--;
            }
            nums[2]--;
            Arrays.sort(nums);
        }
        return res;
    }
}

No.3 构造字典序最大的合并字符串

解题思路

贪心，详见注释。

代码展示

class Solution {
    public String largestMerge(String word1, String word2) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0, j = 0; i < word1.length() || j < word2.length(); ) {
            // word1[i] != word2[j] 时, 选择大的字符
            if (i == word1.length()) {
                sb.append(word2.charAt(j++));
            } else if (j == word2.length()) {
                sb.append(word1.charAt(i++));
            } else if (word1.charAt(i) > word2.charAt(j)) {
                sb.append(word1.charAt(i++));
            } else if (word1.charAt(i) < word2.charAt(j)) {
                sb.append(word2.charAt(j++));
            } else {
                // word1[i] == word2[j] 时
                // 找到第一个不相等的 word1[i + t] != word2[j + t]
                // 选择大的字符所在的字符串
                Character c1 = null, c2 = null;
                for (int t = 1; i + t < word1.length() && j + t < word2.length(); t++) {
                    if (word1.charAt(i + t) != word2.charAt(j + t)) {
                        c1 = word1.charAt(i + t);
                        c2 = word2.charAt(j + t);
                        break;
                    }
                }
                // 若后续全都相等，选择长的
                if (c1 == null && word1.length() - i > word2.length() - j) {
                    sb.append(word1.charAt(i++));
                } else if (c1 == null && word1.length() - i < word2.length() - j) {
                    sb.append(word2.charAt(j++));
                } else if (c1 != null && c1 > c2) {
                    sb.append(word1.charAt(i++));
                } else {
                    sb.append(word2.charAt(j++));
                }
            }
        }
        return sb.toString();
    }
}

No.4 最接近目标值的子序列和

解题思路

超大背包问题。

代码展示

class Solution {
    public int minAbsDifference(int[] nums, int goal) {
        int n = nums.length;
        int half = n / 2;
        int[] a = new int[1 << half];
        for (int i = 0; i < (1 << half); i++) {
            int sum = 0;
            for (int j = 0; j < half; j++) {
                if (((i >> j) & 1) > 0) {
                    sum += nums[j];
                }
            }
            a[i] = sum;
        }
        int half2 = n - half;
        int[] b = new int[1 << half2];
        for (int i = 0; i < (1 << half2); i++) {
            int sum = 0;
            for (int j = 0; j < half2; j++) {
                if (((i >> j) & 1) > 0) {
                    sum += nums[half + j];
                }
            }
            b[i] = sum;
        }
        Arrays.sort(b);
        int ans = Integer.MAX_VALUE;
        for (int v : a) {
            int pos = binarySearch(b, goal - v);
            for (int j = Math.max(0, pos - 2); j < Math.min(b.length, pos + 2); j++) {
                ans = Math.min(ans, Math.abs((v + b[j]) - goal));
            }
        }
        return ans;
    }

    public static int binarySearch(int[] nums, int key) {
        int start = 0, end = nums.length - 1;
        while (start + 1 < end) {
            int mid = (start + end) / 2;
            if (nums[mid] < key) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return nums[start] < key ? end : start;
    }
}

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