LeetCode 算法整理

1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    int i,j,sum;
    int *res = (int *) malloc(sizeof(int)*2);
    for(i=0;i<numsSize;i++){
        res[0] = i;
        for(j=i+1;j<numsSize;j++){ 
            if(nums[i] +nums[j] == target){
                res[1] = j;
                return res;
            }
            
        }
    }
    return res;
}

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    int a=0,b=0,i = 0;
    
    struct ListNode *ln = (struct ListNode*)malloc(sizeof(struct ListNode)); 
    ln->val = 0;
    ln->next = NULL;
    struct ListNode *p = ln;
    struct ListNode *p1 = l1;
    struct ListNode *p2 = l2;
    while(p1!=NULL||p2!=NULL){
        struct ListNode *l = (struct ListNode*)malloc(sizeof(struct ListNode));
        l->val=0;
        l->next = NULL;
        p->next = l;
        p = l; 
        
        if(p1)a=p1->val;else a=0;
        if(p2)b=p2->val;else b=0;
        p->val = a+b+i; 
        if(p->val/10>=1){
            i = 1;
            p->val = p->val%10;
        }else{
            i = 0;
        }
        if(p1)p1=p1->next;
        if(p2)p2=p2->next;
    }
    if(i==1){
        struct ListNode *l = (struct ListNode*)malloc(sizeof(struct ListNode));
        l->val=1;
        l->next = NULL;
        p->next = l;
        p = l; 
    }
    return ln->next;
}

3. Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

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int findstr(char *s,int len,char a){
    int i;
    for(i=0;i<len;i++){
        if(s[i]==a)return 0;
    }
    return 1;
}
int lengthOfLongestSubstring(char* s) {
    int i=0,m=0;
    char a;
    while(*s!='\0'){
        a = s[i];
        while(a!='\0')a=findstr(s,i,a)?s[++i]:'\0';
        if(i>m)m=i;
        s++;i=1;
    }
    return m;
}

4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

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double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    double r = 0;
    int a=0,b=0,i=0,j=0,x=0;
    int *n = (int*)malloc(sizeof(int)*(nums1Size+nums2Size));
    memset(n,9,nums1Size+nums2Size);
    while(x<nums1Size+nums2Size){
        if(i<nums1Size){
            a = nums1[i];
        }else{
            n[x] = nums2[j];
            j++;x++;continue;
        }
        if(j<nums2Size){
            b=nums2[j];
        }else{
            n[x] = nums1[i];
            i++;x++;continue;
        }
        a<b?(i++,n[x]=a):(j++,n[x]=b);
        x++;
    }
    if(x%2!=1)
        r = (n[x/2]+n[x/2-1])/2.0;
    else
        r = n[x/2];
    return r;
}