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Description:
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 andnums2 are m and n respectively.
Code：
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { vector<int> nums(nums1); int itrNums1 = 0, itrNums2 = 0;; vector<int>::iterator itr = nums.begin(); while(itrNums1!=m && itrNums2!=n) { //如果把下面的if判断改为三目运算符会出现"runtime error" if (nums1[itrNums1] <= nums2[itrNums2]) { *itr++ = nums1[itrNums1++]; } else { *itr++ = nums2[itrNums2++]; } } while(itrNums1!=m) { *itr++ = nums1[itrNums1++]; } while(itrNums2!=n) { *itr++ = nums2[itrNums2++]; } for (int i = 0 ; i < m+n; ++i) nums1[i] = nums[i]; }
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原文地址：http://www.cnblogs.com/happygirl-zjj/p/4586271.html
