【LeeCode】485. 最大连续 1 的个数

【题目描述】

给定一个二进制数组 ​​nums​​ ， 计算其中最大连续 ​​1​​ 的个数。

​​​https://leetcode.cn/problems/max-consecutive-ones/​​

【示例】

【代码】admin

package com.company;
// 2022-01-20

class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int count = 0;
int max = 0;
for(int num: nums){
if (num == 1){
count++;
            }else {
max = Math.max(count, max);
count = 0;
            }
        }
max = Math.max(count, max);
System.out.println(max);
return max;
    }
}

public class Test {
public static void main(String[] args) {
new Solution().findMaxConsecutiveOnes(new int[]{1,1,0,1,1,1});  // 输出: 3
new Solution().findMaxConsecutiveOnes(new int[]{1,0,1,1,0,1});  // 输出: 2
    }
}

【代码】​​​​简单​​

package com.company;
// 2022-01-20

class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int max = 0;
int count = 0;

for (int i = 0; i < nums.length; i++){
if (nums[i] == 1){
count++;
            }else {
count = 0;
            }
max = Math.max(max, count);
        }
System.out.println(max);
return max;
    }
}

public class Test {
public static void main(String[] args) {
new Solution().findMaxConsecutiveOnes(new int[]{1,1,0,1,1,1});  // 输出: 3
new Solution().findMaxConsecutiveOnes(new int[]{1,0,1,1,0,1});  // 输出: 2
    }
}

【代码】​​双指针​​

package com.company;
// 2022-01-20

class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int l = 0;
int r = 0;
int n = nums.length;
int res = 0;
while (r < n){
// 获取 连续1的最远下标
while (r < n && nums[r] != 0) r++;
// 始终是r - l的差为最大连续的1
res = Math.max(res, r - l);
// 获取 连续0的最远下标, 此时到到1则停止 
while (r < n && nums[r] != 1) r++;
// 将新的1的位置赋值给l  然后接着统计1最长的小标
l = r;
        }
System.out.println(res);
return res;
    }
}

public class Test {
public static void main(String[] args) {
new Solution().findMaxConsecutiveOnes(new int[]{1,1,0,1,1,1});  // 输出: 3
new Solution().findMaxConsecutiveOnes(new int[]{1,0,1,1,0,1});  // 输出: 2
    }
}