剑指offer专项突击版 ---- 第2天

class Solution {
    public int singleNumber(int[] nums) {
        int ans = 0;
        for(int i = 0; i < 32; i++){
            int target = 0;
            for(int num : nums){
                target += (num >> i) & 1;
            }
            
            ans |= (target%3) << i;
        } 
        return ans;
    }
}

class Solution{
    // 暴力解法
    // m 表示单词的平均长度，n 表示单词的个数
    // 时间复杂度：O(n^2 * m)
    // 空间复杂度：O(1)
    public int maxProduct(String[] words) {
        int ans = 0;
        for (int i = 0; i < words.length -1; i++) {
            String word1 = words[i];
            for (int j = i + 1; j < words.length; j++) {
                String word2 = words[j];
                // 每个单词的 bitMask 会重复计算很多次
                if (!hasSameChar(word1, word2)) {
                    ans = Math.max(ans, word1.length() * word2.length());
                }
            }
        }
        return ans;
    }
    // O(m^2)
    private boolean hasSameChar(String word1, String word2) {
        for (char c : word1.toCharArray()) {
            if (word2.indexOf(c) != -1) return true;
        }
        return false;
    }
}

class Solution {
    // 时间复杂度：O(n)
    // 空间复杂度：O(1)
    public int[] twoSum(int[] nums, int target) {
        if (nums == null || nums.length == 0) return new int[0];

        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            int sum = nums[left] + nums[right];
            if (sum == target) {
                return new int[]{left, right};
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
    return new int[0];
    }
}

标签：专项,right,return,nums,int,offer,----,length,ans
来源： https://blog.csdn.net/DUANJIAWEIDUANJIAWEI/article/details/122135231