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Two pointers因为可能一次要增加两个，所以外层也用while里程while要记录old_j，然后和j比较来确定duplicate的个数 int n = nums.length; int i = 0, j = 0; while (i < n) { int old_j = j; while (j < n && nums[j] == nums[old_j]) { j++; } nums[i++] = nums[old_j]; if (j - old_j > 1) { nums[i++] = nums[old_j]; } if (j == n) { break; } } return i;

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