leetcode485. 最大连续 1 的个数

思路：【双指针】

left左边界，right往右跑遇到0，则计算该长度。并更新cnt（最大连续1个数）。

class Solution {
public:int findMaxConsecutiveOnes(vector<int>& nums) {int left = 0, right = 0;int cnt = 0;while (left != nums.size()) {if (nums[left] == 0) {left++;continue;}right = left;while (right != nums.size()) {if (nums[right] == 0) {int t = right - left;if (t > cnt) {cnt = t;}left = right;break;}right++;}if (right == nums.size()) {int t = right - left;if (t > cnt) {cnt = t;}break;}}return cnt;}
};

 

emmm怎么看答案一次遍历就可以完成呢？

思路：一次遍历

设置两个变量，cnt记录当前小区间的 1 的个数，然后maxcnt记录整个nums数组的 1 的个数（结束一个小区间，满足条件就更新一下）

代码

class Solution {
public:int findMaxConsecutiveOnes(vector<int>& nums) {int count = 0, maxCount = 0;for (int i = 0; i < nums.size(); i++) {if (nums[i] == 1) {count++;}else {maxCount = max(count, maxCount);count = 0;}}maxCount = max(count, maxCount);return maxCount;}
};

 Java 代码

class Solution {public int findMaxConsecutiveOnes(int[] nums) {int cnt = 0, maxcnt = 0;for(int i = 0; i < nums.length; i++){if(nums[i] == 1){cnt++;}else{maxcnt = Math.max(cnt, maxcnt);cnt = 0;}}maxcnt = Math.max(cnt, maxcnt);return maxcnt;}
}