[数据结构与算法]剑指offer 03.53Ⅰ.53Ⅱ. 查找算法(简单)

题目一:

剑指 Offer 03. 数组中重复的数字https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/

想法一:使用flag数组标记是否出现过

想法二:考虑到数字只有0~n-1,所以遍历将每个数字交换到顺序位置,当交换途中位置上数字已经正确时则为重复值

代码:

//法一
class Solution {
    public int findRepeatNumber(int[] nums) {
        boolean[] flag= new boolean[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (flag[nums[i]]){
                return nums[i];
            }
            if (!flag[nums[i]]){
                flag[nums[i]]=true;
            }
        }
        return -1;
    }
}

//法二
class Solution {
    public int findRepeatNumber(int[] nums) {
        int i = 0;
        while(i < nums.length) {
            if(nums[i] == i) {
                i++;
                continue;
            }
            if(nums[nums[i]] == nums[i]) 
                return nums[i];
            int tmp=nums[nums[i]];
            nums[nums[i]]=nums[i];
            nums[i]=tmp;
        }
        return -1;
    }
}

?结果一:

结果二:?

题目二:

剑指 Offer 53 - I. 在排序数组中查找数字 Ihttps://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/

代码:

//法一:遍历计数
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;
    }
}
//法二:二分法查找tar的左右边界
class Solution {
    public int search(int[] nums, int target) {
        return helper(nums, target) - helper(nums, target - 1);
    }
    int helper(int[] nums, int tar) {
        //二分法查找tar右侧数字的位置(该位置可越界)
        //如[5,7,7,8,8,10] 8,tar=8时返回5,tar=7时返回3
        //如[1] 1,tar=1时返回1,tar=0时返回0
        int i = 0, j = nums.length - 1;
        while(i <= j) {
            int m = (i + j) / 2;
            if(nums[m] <= tar) i = m + 1;
            else j = m - 1;
        }
        return i;
    }
}

结果:

?

题目三:

剑指 Offer 53 - II. 0～n-1中缺失的数字https://leetcode-cn.com/problems/que-shi-de-shu-zi-lcof/

代码:

//法一:遍历
class Solution {
    public int missingNumber(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if(nums[i]!=i)
                return i;
        }
        return nums.length;
    }
}
//法二:二分法
class Solution {
    public int missingNumber(int[] nums) {
        int i = 0, j = nums.length - 1;
        while(i <= j) {
            int m = (i + j) / 2;
            if(nums[m] == m) 
                i = m + 1;
            else 
                j = m - 1;
        }
        return i;
    }
}

结果:

?