遍历map,如果有value>1的,则存在,返回true,否则返回false; Runtime:1 ms, faster than96.91%of Java online submissions for Find Subarrays With Equal Sum. Memory Usage:40.1 MB, less than86.76%of Java online submissions for Find Subarrays With Equal Sum....
then find another shortest subarray of sum target whose start index is >= i + 1. This way the two subarrays never overlap. This looks promising. But there is one more issue we need to address: how do we ensure that we are getting the best answer...
A similar idea can be used to partition an array into two subarrays where values from one array are strictly less than values from another array. 1classSolution {2publicstaticvoidmain(String[] args) {3Solution solution =newSolution();4System.out.println(solution.canPartition(newint[]{11, ...