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  • lettCode-Array

    1   Remove Element    lintcode-172

         描述: 删相同元素,反现有长度

         记忆:标不同元素,反标记值

     1  public int removeElement(int[] a, int elem) {
     2         // write your code here
     3         int index = 0;
     4         for (int i = 0; i < a.length; i++) {
     5             if (a[i] != elem) {
     6                 a[index++] = a[i];
     7             }
     8         }
     9         return index;
    10     }
    View Code

     2 Remove Duplicates from Sorted Array 1    lintCode-100

        描述:已排序,删相同,反长度

        记忆:标记不同元素,反标记

     1 public int removeDuplicates(int[] nums) {
     2         // write your code here
     3         if (nums.length == 0) {
     4             return 0;
     5         }
     6         int index = 0;
     7         for (int i = 1; i < nums.length; i++) {
     8             if (nums[index] != nums[i]) {
     9                 nums[++index] = nums[i];
    10             }
    11         }
    12         return index + 1;
    13     }
    View Code

    3 Remove Duplicates from Sorted Array 2    lintCode-101

       描述:已排序,删,最多两,反长度

       记忆:加标记,不等标记少二,才加长

     1     public int removeDuplicates(int[] nums) {
     2         // write your code here
     3         if (nums.length < 2) {
     4             return nums.length;
     5         }
     6         int index = 2;
     7         for (int i = 2; i < nums.length; i++) {
     8             if (nums[i] != nums[index - 2]) {
     9                 nums[index++] = nums[i];
    10             }
    11         }
    12         return index;
    13     } 
    14 }
    View Code

     4 Plus One   lintCode-407

       描述:数组加一反数组

       记忆:加一,进位

     1  public int[] plusOne(int[] digits) {
     2         // Write your code here
     3         int carries = 1;
     4         for (int i = digits.length - 1; i >= 0 && carries > 0; i--) {
     5             int sum = digits[i] + carries;
     6             digits[i] = sum % 10;
     7             carries = sum / 10;
     8         }
     9         
    10         if (carries == 0) {
    11             return digits;
    12         }
    13         
    14         int[] rst = new int[digits.length + 1];
    15         rst[0] = 1;
    16         for (int i = 1; i < rst.length; i++) {
    17             rst[i] = digits[i - 1];
    18         }
    19         
    20         return rst;
    21     }
    View Code

     5 Pascal's Triangle  杨辉三角

      记忆:两数组链表,分类

     1 public static List<Integer> row(int rowIndex) {
     2         List<Integer> res = new ArrayList<Integer>();
     3         int temp = 1;
     4         for (int i = 1; i <= rowIndex; i++) {
     5             List<Integer> list = new ArrayList<Integer>();
     6             if (i == 1) {
     7                 list.add(0, 1);
     8             } else if (i == 2) {
     9                 list.add(0, 1);
    10                 list.add(1, 1);
    11             } else {
    12                 list.add(0, 1);
    13                 for (int k = 0; k < res.size() - 1; k++) {
    14                     int sum = res.get(k) + res.get(k + 1);
    15                     list.add(sum);
    16                 }
    17                 list.add(i - 1, 1);
    18             }
    19             res = list;
    20         }
    21         System.out.println(res);
    22         return res;
    23     }
    View Code
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  • 原文地址:https://www.cnblogs.com/whesuanfa/p/6129317.html
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