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Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.Do not allocate extra space for another array, you must do this in place with constant memory.For example,Given input array nums = [1,1,2],Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length. |
题意：从排好序的数组里删掉重复元素，返回新的数组长度。不能额外申请空间。
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int removeDuplicates(int* nums, int numsSize) {
// int cand=nums[0];
// for(int i=1;i<numsSize;i++){
// if(cand==nums[i]){
// numsSize--;
// }else{
// cand=nums[i];
// }
// }
int index=0;
int j;
for(j=1;j<numsSize;j++){
if(nums[index]!=nums[j]){
nums[++index]=nums[j];
}
}
return index+1;
} |
PS:咦。。。又是一个双指针问题。用两个指针index和j分别指向当前元素和下一个带比较的元素。
index初始为nums[0]，这里一开始我还在想为什么不是从0开始放，其实想错了。。。。。。。。。。。。。。。。。。慢慢悟道吧！！！
本文转自 努力的C 51CTO博客，原文链接:http://blog.51cto.com/fulin0532/1868616