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Commit9430ae6

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add: Target Sum
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‎README.md

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@@ -56,6 +56,7 @@ This is the solutions collection of my LeetCode submissions, most of them are pr
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|404|[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/)|[JavaScript](./src/sum-of-left-leaves/res.js)|Easy|
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|416|[Partition Equal Subset Sum](https://leetcode.com/problems/partition-equal-subset-sum/)|[JavaScript](./src/partition-equal-subset-sum/res.js)|Medium|
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|434|[Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/)|[JavaScript](./src/number-of-segments-in-a-string/res.js)|Easy|
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|494|[Target Sum](https://leetcode.com/problems/target-sum/)|[JavaScript](./src/target-sum/res.js)|Medium|
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|516|[Longest Palindromic Subsequence](https://leetcode.com/problems/longest-palindromic-subsequence/)|[JavaScript](./src/longest-palindromic-subsequence/res.js)|Medium|
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|523|[Continuous Subarray Sum](https://leetcode.com/problems/continuous-subarray-sum/)|[JavaScript](./src/continuous-subarray-sum/res.js)|Medium|
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||[]()|[](./src//res.js)||

‎src/target-sum/res.js

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/**
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* res.js
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*@authors Joe Jiang (hijiangtao@gmail.com)
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*@date 2017-04-12 21:55:49
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*
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* You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.
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*
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* Find out how many ways to assign symbols to make sum of integers equal to target S.
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*
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* Solution idea: https://discuss.leetcode.com/topic/76243/java-15-ms-c-3-ms-o-ns-iterative-dp-solution-using-subset-sum-with-explanation
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*
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*@param {number[]} nums
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*@param {number} S
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*@return {number}
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*/
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letfindTargetSumWays=function(nums,S){
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letlen=nums.length,
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res=0;
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for(leti=0;i<len;i++){
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res+=nums[i];
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}
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if((res<S&&res>-S)||(res+S)&1)return0;
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returncount(nums,len,(res+S)>>>1);
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};
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/**
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* 计算子集存在满足等于 S 的情况数
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*@param {[type]} nums 数组
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*@param {[type]} len 数组长度
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*@param {[type]} S 计算总和
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*@return {[type]} [description]
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*/
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letcount=function(nums,len,S){
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letarr=newArray(S+1);
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arr[0]=1;
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for(leti=1;i<=S;i++){
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arr[i]=0;
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}
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// 遍历 nums 更新 arr 数组
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for(leti=0;i<len;i++){
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for(letj=S;j>=nums[i];j--){
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arr[j]+=arr[j-nums[i]];
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}
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}
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returnarr[S];
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}

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