【Leetcode】485. Max Consecutive Ones
2017-01-16 10:29
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思路:
(1)若数组长度为0,则直接返回0。
(2)count记录当前连续1的个数,max记录到目前为止最大的连续1的个数。
(3)遍历数组,判断每一位,若该位为0,则判断当前连续1的个数是否大于最大值,若是则需要更新最大值,判断完毕后还原count为0;若该位为1,则当前连续1的个数加一。
(4)注意,遍历完毕后需要再判断当前连续1的个数是否大于最大值,若是则需要更新最大值。
(1)若数组长度为0,则直接返回0。
(2)count记录当前连续1的个数,max记录到目前为止最大的连续1的个数。
(3)遍历数组,判断每一位,若该位为0,则判断当前连续1的个数是否大于最大值,若是则需要更新最大值,判断完毕后还原count为0;若该位为1,则当前连续1的个数加一。
(4)注意,遍历完毕后需要再判断当前连续1的个数是否大于最大值,若是则需要更新最大值。
public class Solution { public int findMaxConsecutiveOnes(int[] nums) { int len = nums.length; int count = 0, max = 0; if (len == 0) return 0; for (int i = 0; i < len; i++) { if (nums[i] == 0) { if (count > max) max = count; count = 0; } else count++; } if (count > max) max = count; return max; } }Runtime:7ms
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