Commit 43b9ca1

author

fupengfei

committed

18-01-02 COMMIT

1 parent 06fe836 commit 43b9ca1Copy full SHA for 43b9ca1

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Lines changed: 22 additions & 0 deletions

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	`1`	`+/First Missing Positive:Given an unsorted integer array, find the first missing positive integer./`
	`2`	`+//交换数组元素,使得数组中第i位存放数值(i+1)。最后遍历数组,寻找第一个不符合此要求的元素,返回其下标。`
	`3`	`+class Solution {`
	`4`	`+public:`
	`5`	`+ int firstMissingPositive(vector<int>& nums) {`
	`6`	`+ int i = 0;`
	`7`	`+ int n = nums.size();`
	`8`	`+ while(i < n){`
	`9`	`+ if(nums[i] != i+1 && nums[i] >= 1 && nums[i] <= n && nums[nums[i]-1] != nums[i]){`
	`10`	`+ swap(nums[i], nums[nums[i]-1]);`
	`11`	`+ }else{`
	`12`	`+ i++;`
	`13`	`+ }`
	`14`	`+ }`
	`15`	`+ for(i = 0; i < n; ++i){`
	`16`	`+ if(nums[i] != i+1){`
	`17`	`+ return i+1;`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+ return n+1;`
	`21`	`+ }`
	`22`	`+};`

Lines changed: 31 additions & 0 deletions

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	`1`	`+/*Trapping Rain Water:Given n non-negative integers representing an elevation map where the width of each bar is 1,`
	`2`	`+compute how much water it is able to trap after raining.*/`
	`3`	`+class Solution {`
	`4`	`+public:`
	`5`	`+ int trap(vector<int>& height) {`
	`6`	`+ if(height.size() < 3) return 0;`
	`7`	`+ int left = 0;`
	`8`	`+ int right = height.size() - 1;`
	`9`	`+ int sum = 0;`
	`10`	`+ // 找到左边的第一个峰值`
	`11`	`+ while(left < right && height[left] <= height[left+1]) left++;`
	`12`	`+ // 找到右边的第一个峰值`
	`13`	`+ while(left < right && height[right] <= height[right-1]) right--;`
	`14`	`+ while(left < right){`
	`15`	`+ int leftVal = height[left];`
	`16`	`+ int rightVal = height[right];`
	`17`	`+ // 如果左边峰值较小,先计算左边`
	`18`	`+ if(leftVal < rightVal){`
	`19`	`+ while(left < right && leftVal >= height[++left]){`
	`20`	`+ sum += leftVal - height[left];`
	`21`	`+ }`
	`22`	`+ // 如果右边峰值较小,先计算右边`
	`23`	`+ } else {`
	`24`	`+ while(left < right && rightVal >= height[--right]){`
	`25`	`+ sum += rightVal - height[right];`
	`26`	`+ }`
	`27`	`+ }`
	`28`	`+ }`
	`29`	`+ return sum;`
	`30`	`+ }`
	`31`	`+};`

Lines changed: 35 additions & 0 deletions

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	`1`	`+/Multiply Strings:Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2./`
	`2`	`+class Solution {`
	`3`	`+public:`
	`4`	`+ string multiply(string num1, string num2) {`
	`5`	`+ static const int R = 10;`
	`6`	`+`
	`7`	`+ int n1 = num1.size();`
	`8`	`+ int n2 = num2.size();`
	`9`	`+ int i;`
	`10`	`+ int j;`
	`11`	`+ string num3 = "";`
	`12`	`+ int sum;`
	`13`	`+ int carry;`
	`14`	`+ int low, high;`
	`15`	`+`
	`16`	`+ carry = 0;`
	`17`	`+ for (i = 0; i < n1 + n2 - 1; ++i) {`
	`18`	`+ low = i > n2 - 1 ? i - (n2 - 1) : 0;`
	`19`	`+ high = i < n1 - 1 ? i : n1 - 1;`
	`20`	`+ sum = carry;`
	`21`	`+ for (j = low; j <= high; ++j) {`
	`22`	`+ sum += (num1[n1 - 1 - j] - '0') * (num2[n2 - 1 - (i - j)] - '0');`
	`23`	`+ }`
	`24`	`+ carry = sum / R;`
	`25`	`+ sum %= R;`
	`26`	`+ num3.push_back(sum + '0');`
	`27`	`+ }`
	`28`	`+ num3.push_back(carry + '0');`
	`29`	`+ while (num3.size() > 1 && num3.back() == '0') {`
	`30`	`+ num3.pop_back();`
	`31`	`+ }`
	`32`	`+ reverse(num3.begin(), num3.end());`
	`33`	`+ return num3;`
	`34`	`+ }`
	`35`	`+};`

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