LeetCode : 485. Max Consecutive Ones
2017-02-21 21:45
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代码:
int findMaxConsecutiveOnes(int* nums, int numsSize) {
int i;
int tag = 0;
int max = 0;
for(i = 0;i < numsSize;i++)
{
if(nums[i] == 1){
tag++;
if(tag > max)
{
max = tag;
}
}
else {tag = 0;}
}
return max;
}
简要题解:
(1)设立一个标签tag,一个最大值max,两者的初始值均为0;
(2)一次遍历输入的数组,如果当前数组字符为1,tag加1,如果tag比max大,则max的值就是tag的值,如果当前数组字符为0,tag为0;
(3)返回max的值
int findMaxConsecutiveOnes(int* nums, int numsSize) {
int i;
int tag = 0;
int max = 0;
for(i = 0;i < numsSize;i++)
{
if(nums[i] == 1){
tag++;
if(tag > max)
{
max = tag;
}
}
else {tag = 0;}
}
return max;
}
简要题解:
(1)设立一个标签tag,一个最大值max,两者的初始值均为0;
(2)一次遍历输入的数组,如果当前数组字符为1,tag加1,如果tag比max大,则max的值就是tag的值,如果当前数组字符为0,tag为0;
(3)返回max的值
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