【LeetCode】485. Max Consecutive Ones

问题描述

问题链接：https://leetcode.com/problems/max-consecutive-ones/#/description

Given a binary array, find the maximum number of consecutive 1s in this array.

Example 1:

Input: [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s.
The maximum number of consecutive 1s is 3.

Note:

The input array will only contain 0 and 1.

The length of input array is a positive integer and will not exceed 10,000

我的代码

思路比较简单，直接上代码了。

public class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
/*
思路是遍历整个数组，如果碰到1，curLen ++。然后继续向后
如果碰到了0，比较一下curLen和maxLen,如果curLen>maxLen,则maxLen = curLen
*/

int maxLen = 0;
int curLen = 0;

int len = nums.length;
for(int i = 0; i < len; i++){
if(nums[i] == 1){
curLen ++;
}
if(nums[i] == 0 || i == len - 1){
if(curLen > maxLen){
maxLen = curLen;
}
curLen = 0;
}
}
return maxLen;
}
}

打败了55.28%的Java代码，来看看讨论区的大神们。

讨论区

Java 4 lines concise solution with explanation

链接地址：https://discuss.leetcode.com/topic/75437/java-4-lines-concise-solution-with-explanation

这个蛮简洁的。

public int findMaxConsecutiveOnes(int[] nums) {
int maxHere = 0, max = 0;
for (int n : nums)
max = Math.max(max, maxHere = n == 0 ? 0 : maxHere + 1);
return max;
}

Easy Java Solution

链接地址：https://discuss.leetcode.com/topic/75430/easy-java-solution

感觉这个就是我的加强版，就是要这种感觉！

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

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

return result;
}
}