leetcode 674. 无序整数数组，找最长且连续递增的子序列

末蓝、 2024-03-27 17:06 52阅读 0赞

> [牛客华为机试题库【题号 HJ开头】（重点看）][HJ]  
> [牛客在线编程算法篇【题号NC开头】][NC]  
> [剑指offer【题号 JZ开头】][offer_ JZ]  
> [力扣][Link 1]

### 1）[原题链接][Link 2] ###

### 2）已有题解 ###

### 3）代码 ###

package doublepoint;
    
    import java.io.BufferedReader;
    import java.io.IOException;
    import java.io.InputStreamReader;
    import java.util.Arrays;
    
    import static java.lang.System.in;
    
    /**
     * leetcode 674. 无序整数数组的最长且连续递增的子序列
     */
    public class leetcode674LongestIncreasingContinuousSubsequence {
        
        public static void main(String[] args) throws IOException {
        
            BufferedReader bf = new BufferedReader(new InputStreamReader(in));
            int[] ints = Arrays.stream(bf.readLine().split(",")).mapToInt(Integer::parseInt).toArray();
            bf.close();
    
            int lengthOfLCIS = findLengthOfLCIS(ints);
            System.out.println(lengthOfLCIS);
    
            int[] lengthOfLCIS1 = findLengthOfLCIS1(ints);
            System.out.println(Arrays.toString(lengthOfLCIS1));
        }
    
        /**
         * leetcode 674. 无序整数数组的最长且连续递增的子序列的长度
         */
        public static int findLengthOfLCIS(int[] nums) {
        
            if (nums.length == 0) {
        
                return 0;
            }
    
            if (nums.length == 1) {
        
                return 1;
            }
    
            int max = 1;
    
            int start = 0;
            for (int i = 1; i < nums.length; i++) {
        
                if (nums[i - 1] < nums[i]) {
        
                    // 5-3=2表示5比3大2个，3到5一共有三个，故再加1
                    max = Math.max(i - start + 1, max);
                } else {
        
                    start = i;
                }
            }
    
            return max;
        }
    
        /**
         * leetcode 674. 无序整数数组的最长且连续递增的子序列
         */
        public static int[] findLengthOfLCIS1(int[] nums) {
        
            if (nums.length < 2) {
         // 2个一下不涉及排序
                return nums;
            }
    
            int max = 1;
            int[] temp = {
        nums[0]};  // 下边从1开始和前面的值比较,起码有前面这个值,所以收集第一个元素应对[2,2,2,2,2]这种场景
    
            int start = 0;
            for (int i = 1; i < nums.length; i++) {
        
                if (nums[i - 1] < nums[i]) {
        
                    if (i - start + 1 > max) {
        
                        max = i - start + 1;
                        // 方法1：直接截取数组
                        temp = Arrays.copyOfRange(nums, start, i + 1);
                        // 方法2：记录开始结束坐标最后再截取
                    }
                } else {
        
                    start = i;
                }
            }
    
            return temp;
        }
    }

[HJ]: https://www.nowcoder.com/exam/oj/ta?tpId=37
[NC]: https://www.nowcoder.com/exam/oj
[offer_ JZ]: https://www.nowcoder.com/exam/oj/ta?page=1&tpId=13&type=13
[Link 1]: https://leetcode.cn/problemset/all/
[Link 2]: https://leetcode.cn/problems/longest-continuous-increasing-subsequence/